blocks|key|3952635|text|我是这样做的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3952636|private+string+GetExcelColumnName(int+columnNumber)
{
++++string+columnName+=+"";

++++while+(columnNumber+>+0)
++++{
++++++++int+modulo+=+(columnNumber+-+1)+%25+26;
++++++++columnName+=+Convert.ToChar('A'+%2B+modulo)+%2B+columnName;
++++++++columnNumber+=+(columnNumber+-+modulo)+/+26;
++++}+

++++return+columnName;
}|code-block|syntax|javascript|3952637|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's how I do it:
<pre><code>private string GetExcelColumnName(int columnNumber)
{
 string columnName = &quot;&quot;;

 while (columnNumber &gt; 0)
 {
 int modulo = (columnNumber - 1) % 26;
 columnName = Convert.ToChar('A' + modulo) + columnName;
 columnNumber = (columnNumber - modulo) / 26;
 } 

 return columnName;
}
</code></pre>

blocks|key|810687|text|如果任何人需要在没有VBA的情况下在Excel中执行此操作，这里有一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810688|=SUBSTITUTE(ADDRESS(1;colNum;4);"1";"")|code-block|syntax|javascript|810689|其中colNum是列号|810690|在VBA中：|810691|Function+GetColumnName(colNum+As+Integer)+As+String
++++Dim+d+As+Integer
++++Dim+m+As+Integer
++++Dim+name+As+String
++++d+=+colNum
++++name+=+""
++++Do+While+(d+>+0)
++++++++m+=+(d+-+1)+Mod+26
++++++++name+=+Chr(65+%2B+m)+%2B+name
++++++++d+=+Int((d+-+m)+/+26)
++++Loop
++++GetColumnName+=+name
End+Function|810692|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If anyone needs to do this in Excel without VBA, here is a way:

<pre><code>=SUBSTITUTE(ADDRESS(1;colNum;4);"1";"")
</code></pre>

where colNum is the column number

And in VBA:

<pre><code>Function GetColumnName(colNum As Integer) As String
 Dim d As Integer
 Dim m As Integer
 Dim name As String
 d = colNum
 name = ""
 Do While (d &gt; 0)
 m = (d - 1) Mod 26
 name = Chr(65 + m) + name
 d = Int((d - m) / 26)
 Loop
 GetColumnName = name
End Function
</code></pre>

blocks|key|3106962|text|对不起，这是Python而不是C#，但至少结果是正确的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3106963|def+ColIdxToXlName(idx):
++++if+idx+<+1:
++++++++raise+ValueError("Index+is+too+small")
++++result+=+""
++++while+True:
++++++++if+idx+>+26:
++++++++++++idx,+r+=+divmod(idx+-+1,+26)
++++++++++++result+=+chr(r+%2B+ord('A'))+%2B+result
++++++++else:
++++++++++++return+chr(idx+%2B+ord('A')+-+1)+%2B+result


for+i+in+xrange(1,+1024):
++++print+"%254d+:+%25s"+%25+(i,+ColIdxToXlName(i))|code-block|syntax|javascript|3106964|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Sorry, this is Python instead of C#, but at least the results are correct:

<pre><code>def ColIdxToXlName(idx):
 if idx &lt; 1:
 raise ValueError("Index is too small")
 result = ""
 while True:
 if idx &gt; 26:
 idx, r = divmod(idx - 1, 26)
 result = chr(r + ord('A')) + result
 else:
 return chr(idx + ord('A') - 1) + result


for i in xrange(1, 1024):
 print "%4d : %s" % (i, ColIdxToXlName(i))
</code></pre>

blocks|key|806395|text|您可能需要两种方式的转换，例如，从像AAZ这样的Excel列地址转换为整数，或者从任何整数转换为Excel。下面的两个方法就可以做到这一点。假设基于1的索引，您的“数组”中的第一个元素是元素编号1。这里没有大小限制，所以您可以使用诸如ERROR之类的地址，这将是列编号2613824+...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|806396|public+static+string+ColumnAdress(int+col)
{
++if+(col+<=+26)+{+
++++return+Convert.ToChar(col+%2B+64).ToString();
++}
++int+div+=+col+/+26;
++int+mod+=+col+%25+26;
++if+(mod+==+0)+{mod+=+26;div--;}
++return+ColumnAdress(div)+%2B+ColumnAdress(mod);
}

public+static+int+ColumnNumber(string+colAdress)
{
++int[]+digits+=+new+int[colAdress.Length];
++for+(int+i+=+0;+i+<+colAdress.Length;+%2B%2Bi)
++{
++++digits[i]+=+Convert.ToInt32(colAdress[i])+-+64;
++}
++int+mul=1;int+res=0;
++for+(int+pos+=+digits.Length+-+1;+pos+>=+0;+--pos)
++{
++++res+%2B=+digits[pos]+*+mul;
++++mul+*=+26;
++}
++return+res;
}|code-block|syntax|javascript|806397|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You might need conversion both ways, e.g from Excel column adress like AAZ to integer and from any integer to Excel. The two methods below will do just that. Assumes 1 based indexing, first element in your "arrays" are element number 1.
No limits on size here, so you can use adresses like ERROR and that would be column number 2613824 ...

<pre><code>public static string ColumnAdress(int col)
{
 if (col &lt;= 26) { 
 return Convert.ToChar(col + 64).ToString();
 }
 int div = col / 26;
 int mod = col % 26;
 if (mod == 0) {mod = 26;div--;}
 return ColumnAdress(div) + ColumnAdress(mod);
}

public static int ColumnNumber(string colAdress)
{
 int[] digits = new int[colAdress.Length];
 for (int i = 0; i &lt; colAdress.Length; ++i)
 {
 digits[i] = Convert.ToInt32(colAdress[i]) - 64;
 }
 int mul=1;int res=0;
 for (int pos = digits.Length - 1; pos &gt;= 0; --pos)
 {
 res += digits[pos] * mul;
 mul *= 26;
 }
 return res;
}
</code></pre>

blocks|key|3107489|text|我在我的第一篇文章中发现了一个错误，所以我决定坐下来计算一下。我发现用来识别Excel列的数字系统并不是一个基数为26的系统，正如另一个人发布的那样。考虑以10为基数的以下内容，也可以使用字母表中的字母。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3107490|Space:.........................S1，S2，S3+:+S1，S2，S3|3107491|....................................0，00,000+:.A、AA、AAA|3107492|....................................1，01,001+:.B、AB、AAB|3107493|..…，…，…:..…，…，…|3107494|....................................9，99,999+:..Z，ZZ，ZZZ|3107495|空间状态总数:101001000:26676,17576|3107496|总States:...............1110................18278|3107497|Excel使用基数26对各个字母空格中的列进行编号。您可以看到，一般而言，状态空间级数是a，a%5E2，a%5E3，…对于某些基数a，状态总数是a%2B+a%5E2+%2B+a%5E3+%2B…。|3107498|假设您想要找到前N个空间中的状态A的总数。计算公式为A=+(a)(a%5EN+-1+)/(a-1)。这很重要，因为我们需要找到对应于索引K的空间N。如果我想找出K在数字系统中的位置，我需要用K替换A并求解N。解决方案是N=+log{base+a}+(A+(a-1)/a+%2B1)。如果我使用a=+10和K=+192的例子，我知道N=+2.23804…。这告诉我，K位于第三个空间的开头，因为它略大于2。|3107499|下一步是找出我们在当前空间中到底有多远。要找到这一点，从K中减去使用N的楼层生成的A。在本例中，N的楼层是2。因此，A=+(10)(10%5E2+-+1)/(10-1)+=+110，正如您在组合前两个空间的状态时所预期的那样。这需要从K中减去，因为前110个状态已经在前两个空间中被考虑到了。这就只剩下82个州了。因此，在这个数字系统中，以10为基数的192的表示是082。|3107500|使用基础索引0的C#代码是|3107501|++++private+string+ExcelColumnIndexToName(int+Index)
++++{
++++++++string+range+=+string.Empty;
++++++++if+(Index+<+0+)+return+range;
++++++++int+a+=+26;
++++++++int+x+=+(int)Math.Floor(Math.Log((Index)+*+(a+-+1)+/+a+%2B+1,+a));
++++++++Index+-=+(int)(Math.Pow(a,+x)+-+1)+*+a+/+(a+-+1);
++++++++for+(int+i+=+x%2B1;+Index+%2B+i+>+0;+i--)
++++++++{
++++++++++++range+=+((char)(65+%2B+Index+%25+a)).ToString()+%2B+range;
++++++++++++Index+/=+a;
++++++++}
++++++++return+range;
++++}|code-block|syntax|javascript|3107502|//旧帖子|3107503|C#中从零开始的解决方案。|3107504|++++private+string+ExcelColumnIndexToName(int+Index)
++++{
++++++++string+range+=+"";
++++++++if+(Index+<+0+)+return+range;
++++++++for(int+i=1;Index+%2B+i+>+0;i=0)
++++++++{
++++++++++++range+=+((char)(65+%2B+Index+%25+26)).ToString()+%2B+range;
++++++++++++Index+/=+26;
++++++++}
++++++++if+(range.Length+>+1)+range+=+((char)((int)range[0]+-+1)).ToString()+%2B+range.Substring(1);
++++++++return+range;
++++}|3107505|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1C|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1D|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|1E|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1F|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1G|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1H|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1I|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1J|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1K|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1L|8|@]|9|@]|A|$]]|$1|X|3|Y|5|Z|7|1M|8|@]|9|@]|A|$10|11]]|$1|12|3|13|5|6|7|1N|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|1O|8|@]|9|@]|A|$]]|$1|16|3|17|5|Z|7|1P|8|@]|9|@]|A|$10|11]]|$1|18|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]]|19|$]]

I discovered an error in my first post, so I decided to sit down and do the the math. What I found is that the number system used to identify Excel columns is not a base 26 system, as another person posted. Consider the following in base 10. You can also do this with the letters of the alphabet.

Space:.........................S1, S2, S3 : S1, S2, S3 
....................................0, 00, 000 :.. A, AA, AAA 
....................................1, 01, 001 :.. B, AB, AAB 
.................................... …, …, … :.. …, …, … 
....................................9, 99, 999 :.. Z, ZZ, ZZZ 
Total states in space: 10, 100, 1000 : 26, 676, 17576 
Total States:...............1110................18278

Excel numbers columns in the individual alphabetical spaces using base 26. You can see that in general, the state space progression is a, a^2, a^3, … for some base a, and the total number of states is a + a^2 + a^3 + … . 

Suppose you want to find the total number of states A in the first N spaces. The formula for doing so is A = (a)(a^N - 1 )/(a-1). This is important because we need to find the space N that corresponds to our index K. If I want to find out where K lies in the number system I need to replace A with K and solve for N. The solution is N = log{base a} (A (a-1)/a +1). If I use the example of a = 10 and K = 192, I know that N = 2.23804… . This tells me that K lies at the beginning of the third space since it is a little greater than two.

The next step is to find exactly how far in the current space we are. To find this, subtract from K the A generated using the floor of N. In this example, the floor of N is two. So, A = (10)(10^2 – 1)/(10-1) = 110, as is expected when you combine the states of the first two spaces. This needs to be subtracted from K because these first 110 states would have already been accounted for in the first two spaces. This leaves us with 82 states. So, in this number system, the representation of 192 in base 10 is 082. 

The C# code using a base index of zero is

<pre><code> private string ExcelColumnIndexToName(int Index)
 {
 string range = string.Empty;
 if (Index &lt; 0 ) return range;
 int a = 26;
 int x = (int)Math.Floor(Math.Log((Index) * (a - 1) / a + 1, a));
 Index -= (int)(Math.Pow(a, x) - 1) * a / (a - 1);
 for (int i = x+1; Index + i &gt; 0; i--)
 {
 range = ((char)(65 + Index % a)).ToString() + range;
 Index /= a;
 }
 return range;
 }
</code></pre>

//Old Post

A zero-based solution in C#.

<pre><code> private string ExcelColumnIndexToName(int Index)
 {
 string range = "";
 if (Index &lt; 0 ) return range;
 for(int i=1;Index + i &gt; 0;i=0)
 {
 range = ((char)(65 + Index % 26)).ToString() + range;
 Index /= 26;
 }
 if (range.Length &gt; 1) range = ((char)((int)range[0] - 1)).ToString() + range.Substring(1);
 return range;
 }
</code></pre>

blocks|key|3106895|text|使用递归使变得简单。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3106896|public+static+string+GetStandardExcelColumnName(int+columnNumberOneBased)
{
++int+baseValue+=+Convert.ToInt32('A');
++int+columnNumberZeroBased+=+columnNumberOneBased+-+1;

++string+ret+=+"";

++if+(columnNumberOneBased+>+26)
++{
++++ret+=+GetStandardExcelColumnName(columnNumberZeroBased+/+26)+;
++}

++return+ret+%2B+Convert.ToChar(baseValue+%2B+(columnNumberZeroBased+%25+26)+);
}|code-block|syntax|javascript|3106897|entityMap^0|5|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Easy with recursion.

<pre><code>public static string GetStandardExcelColumnName(int columnNumberOneBased)
{
 int baseValue = Convert.ToInt32('A');
 int columnNumberZeroBased = columnNumberOneBased - 1;

 string ret = "";

 if (columnNumberOneBased &gt; 26)
 {
 ret = GetStandardExcelColumnName(columnNumberZeroBased / 26) ;
 }

 return ret + Convert.ToChar(baseValue + (columnNumberZeroBased % 26) );
}
</code></pre>

blocks|key|806687|text|答案在javaScript中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|806688|function+getCharFromNumber(columnNumber){
++++var+dividend+=+columnNumber;
++++var+columnName+=+"";
++++var+modulo;

++++while+(dividend+>+0)
++++{
++++++++modulo+=+(dividend+-+1)+%25+26;
++++++++columnName+=+String.fromCharCode(65+%2B+modulo).toString()+%2B+columnName;
++++++++dividend+=+parseInt((dividend+-+modulo)+/+26);
++++}+
++++return++columnName;
}|code-block|syntax|javascript|806689|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This answer is in javaScript:

<pre><code>function getCharFromNumber(columnNumber){
 var dividend = columnNumber;
 var columnName = "";
 var modulo;

 while (dividend &gt; 0)
 {
 modulo = (dividend - 1) % 26;
 columnName = String.fromCharCode(65 + modulo).toString() + columnName;
 dividend = parseInt((dividend - modulo) / 26);
 } 
 return columnName;
}
</code></pre>

blocks|key|806138|text|int+nCol+=+127;
string+sChars+=+"ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string+sCol+=+"";
while+(nCol+>=+26)
{
++++int+nChar+=+nCol+%25+26;
++++nCol+=+(nCol+-+nChar)+/+26;
++++//+You+could+do+some+trick+with+using+nChar+as+offset+from+'A',+but+I+am+lazy+to+do+it+right+now.
++++sCol+=+sChars[nChar]+%2B+sCol;
}
sCol+=+sChars[nCol]+%2B+sCol;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|806139|更新：Peter的评论是正确的。这就是我在浏览器中编写代码得到的结果。:-)我的解决方案没有编译，它缺少最左边的字母，它以相反的顺序构建字符串-所有问题现在都解决了。|unstyled|offset|length|style|BOLD|806140|抛开Bugs不谈，该算法基本上是将一个数字从基数10转换为基数26。|806141|更新2：Joel+Coehoorn是对的-上面的代码将返回27的AB。如果是以26为底的实数，AA将等于A，Z之后的下一个数将是BA。|806142|int+nCol+=+127;
string+sChars+=+"0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string+sCol+=+"";
while+(nCol+>+26)
{
++++int+nChar+=+nCol+%25+26;
++++if+(nChar+==+0)
++++++++nChar+=+26;
++++nCol+=+(nCol+-+nChar)+/+26;
++++sCol+=+sChars[nChar]+%2B+sCol;
}
if+(nCol+!=+0)
++++sCol+=+sChars[nCol]+%2B+sCol;|806143|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/5189/peter|1|https://stackoverflow.com/users/3043/joel-coehoorn^0|0|0|2|3|5|0|0|0|0|3|4|D|1|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|11|8|@$G|12|H|13|I|J]]|9|@$G|14|H|15|1|16]]|A|$]]|$1|K|3|L|5|F|7|17|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|18|8|@$G|19|H|1A|I|J]]|9|@$G|1B|H|1C|1|1D]]|A|$]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$B|C]]|$1|Q|3|-4|5|F|7|1F|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

<pre><code>int nCol = 127;
string sChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol &gt;= 26)
{
 int nChar = nCol % 26;
 nCol = (nCol - nChar) / 26;
 // You could do some trick with using nChar as offset from 'A', but I am lazy to do it right now.
 sCol = sChars[nChar] + sCol;
}
sCol = sChars[nCol] + sCol;
</code></pre>

Update: <a href="https://stackoverflow.com/users/5189/peter">Peter</a>'s comment is right. That's what I get for writing code in the browser. :-) My solution was not compiling, it was missing the left-most letter and it was building the string in reverse order - all now fixed.

Bugs aside, the algorithm is basically converting a number from base 10 to base 26.

Update 2: <a href="https://stackoverflow.com/users/3043/joel-coehoorn">Joel Coehoorn</a> is right - the code above will return AB for 27. If it was real base 26 number, AA would be equal to A and the next number after Z would be BA.

<pre><code>int nCol = 127;
string sChars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol &gt; 26)
{
 int nChar = nCol % 26;
 if (nChar == 0)
 nChar = 26;
 nCol = (nCol - nChar) / 26;
 sCol = sChars[nChar] + sCol;
}
if (nCol != 0)
 sCol = sChars[nCol] + sCol;
</code></pre>

blocks|key|3953507|text|令我惊讶的是，到目前为止所有的解决方案都包含迭代或递归。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3953508|这是我的解决方案，在固定的时间内运行(没有循环)。此解决方案适用于所有可能的Excel列，并检查输入是否可以转换为Excel列。可能的列在A，XFD或1,16384的范围内。(这取决于您的Excel版本)|3953509|private+static+string+Turn(uint+col)
{
++++if+(col+<+1+%7C%7C+col+>+16384)+//Excel+columns+are+one-based+(one+=+'A')
++++++++throw+new+ArgumentException("col+must+be+>=+1+and+<=+16384");

++++if+(col+<=+26)+//one+character
++++++++return+((char)(col+%2B+'A'+-+1)).ToString();

++++else+if+(col+<=+702)+//two+characters
++++{
++++++++char+firstChar+=+(char)((int)((col+-+1)+/+26)+%2B+'A'+-+1);
++++++++char+secondChar+=+(char)(col+%25+26+%2B+'A'+-+1);

++++++++if+(secondChar+==+'@')+//Excel+is+one-based,+but+modulo+operations+are+zero-based
++++++++++++secondChar+=+'Z';+//convert+one-based+to+zero-based

++++++++return+string.Format("{0}{1}",+firstChar,+secondChar);
++++}

++++else+//three+characters
++++{
++++++++char+firstChar+=+(char)((int)((col+-+1)+/+702)+%2B+'A'+-+1);
++++++++char+secondChar+=+(char)((col+-+1)+/+26+%25+26+%2B+'A'+-+1);
++++++++char+thirdChar+=+(char)(col+%25+26+%2B+'A'+-+1);

++++++++if+(thirdChar+==+'@')+//Excel+is+one-based,+but+modulo+operations+are+zero-based
++++++++++++thirdChar+=+'Z';+//convert+one-based+to+zero-based

++++++++return+string.Format("{0}{1}{2}",+firstChar,+secondChar,+thirdChar);
++++}
}|code-block|syntax|javascript|3953510|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I'm surprised all of the solutions so far contain either iteration or recursion.

Here's my solution that runs in constant time (no loops). This solution works for all possible Excel columns and checks that the input can be turned into an Excel column. Possible columns are in the range [A, XFD] or [1, 16384]. (This is dependent on your version of Excel)

<pre><code>private static string Turn(uint col)
{
 if (col &lt; 1 || col &gt; 16384) //Excel columns are one-based (one = 'A')
 throw new ArgumentException("col must be &gt;= 1 and &lt;= 16384");

 if (col &lt;= 26) //one character
 return ((char)(col + 'A' - 1)).ToString();

 else if (col &lt;= 702) //two characters
 {
 char firstChar = (char)((int)((col - 1) / 26) + 'A' - 1);
 char secondChar = (char)(col % 26 + 'A' - 1);

 if (secondChar == '@') //Excel is one-based, but modulo operations are zero-based
 secondChar = 'Z'; //convert one-based to zero-based

 return string.Format("{0}{1}", firstChar, secondChar);
 }

 else //three characters
 {
 char firstChar = (char)((int)((col - 1) / 702) + 'A' - 1);
 char secondChar = (char)((col - 1) / 26 % 26 + 'A' - 1);
 char thirdChar = (char)(col % 26 + 'A' - 1);

 if (thirdChar == '@') //Excel is one-based, but modulo operations are zero-based
 thirdChar = 'Z'; //convert one-based to zero-based

 return string.Format("{0}{1}{2}", firstChar, secondChar, thirdChar);
 }
}
</code></pre>

blocks|key|3954212|text|只是使用递归抛出了一个简单的两行C#实现，因为这里的所有答案似乎都比必要的复杂得多。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3954213|///+<summary>
///+Gets+the+column+letter(s)+corresponding+to+the+given+column+number.
///+</summary>
///+<param+name="column">The+one-based+column+index.+Must+be+greater+than+zero.</param>
///+<returns>The+desired+column+letter,+or+an+empty+string+if+the+column+number+was+invalid.</returns>
public+static+string+GetColumnLetter(int+column)+{
++++if+(column+<+1)+return+String.Empty;
++++return+GetColumnLetter((column+-+1)+/+26)+%2B+(char)('A'+%2B+(column+-+1)+%25+26);
}|code-block|syntax|javascript|3954214|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just throwing in a simple two-line C# implementation using recursion, because all the answers here seem far more complicated than necessary.

<pre><code>/// &lt;summary&gt;
/// Gets the column letter(s) corresponding to the given column number.
/// &lt;/summary&gt;
/// &lt;param name="column"&gt;The one-based column index. Must be greater than zero.&lt;/param&gt;
/// &lt;returns&gt;The desired column letter, or an empty string if the column number was invalid.&lt;/returns&gt;
public static string GetColumnLetter(int column) {
 if (column &lt; 1) return String.Empty;
 return GetColumnLetter((column - 1) / 26) + (char)('A' + (column - 1) % 26);
}
</code></pre>

blocks|key|810778|text|..And转换为php：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810779|function+GetExcelColumnName($columnNumber)+{
++++$columnName+=+'';
++++while+($columnNumber+>+0)+{
++++++++$modulo+=+($columnNumber+-+1)+%25+26;
++++++++$columnName+=+chr(65+%2B+$modulo)+.+$columnName;
++++++++$columnNumber+=+(int)(($columnNumber+-+$modulo)+/+26);
++++}
++++return+$columnName;
}|code-block|syntax|javascript|810780|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

..And converted to php:

<pre><code>function GetExcelColumnName($columnNumber) {
 $columnName = '';
 while ($columnNumber &gt; 0) {
 $modulo = ($columnNumber - 1) % 26;
 $columnName = chr(65 + $modulo) . $columnName;
 $columnNumber = (int)(($columnNumber - $modulo) / 26);
 }
 return $columnName;
}
</code></pre>

blocks|key|3953186|text|Java中的相同实现|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3953187|public+String+getExcelColumnName+(int+columnNumber)+
++++{+++++
++++++++int+dividend+=+columnNumber;+++
++++++++int+i;
++++++++String+columnName+=+"";+++++
++++++++int+modulo;+++++
++++++++while+(dividend+>+0)+++++
++++++++{++++++++
++++++++++++modulo+=+(dividend+-+1)+%25+26;+++++++++
++++++++++++i+=+65+%2B+modulo;
++++++++++++columnName+=+new+Character((char)i).toString()+%2B+columnName;++++++++
++++++++++++dividend+=+(int)((dividend+-+modulo)+/+26);++++
++++++++}+++++++
++++++++return+columnName;+
++++}++|code-block|syntax|javascript|3953188|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Same implementation in Java
<pre><code>public String getExcelColumnName (int columnNumber) 
 { 
 int dividend = columnNumber; 
 int i;
 String columnName = &quot;&quot;; 
 int modulo; 
 while (dividend &gt; 0) 
 { 
 modulo = (dividend - 1) % 26; 
 i = 65 + modulo;
 columnName = new Character((char)i).toString() + columnName; 
 dividend = (int)((dividend - modulo) / 26); 
 } 
 return columnName; 
 } 
</code></pre>

blocks|key|3953764|text|我想抛出我使用的静态类，用于在col和col之间进行交互。我对我的ColumnLabel方法使用了修改后的接受答案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3953765|public+static+class+Extensions
{
++++public+static+string+ColumnLabel(this+int+col)
++++{
++++++++var+dividend+=+col;
++++++++var+columnLabel+=+string.Empty;
++++++++int+modulo;

++++++++while+(dividend+>+0)
++++++++{
++++++++++++modulo+=+(dividend+-+1)+%25+26;
++++++++++++columnLabel+=+Convert.ToChar(65+%2B+modulo).ToString()+%2B+columnLabel;
++++++++++++dividend+=+(int)((dividend+-+modulo)+/+26);
++++++++}+

++++++++return+columnLabel;
++++}
++++public+static+int+ColumnIndex(this+string+colLabel)
++++{
++++++++//+"AD"+(1+*+26%5E1)+%2B+(4+*+26%5E0)+...
++++++++var+colIndex+=+0;
++++++++for(int+ind+=+0,+pow+=+colLabel.Count()-1;+ind+<+colLabel.Count();+%2B%2Bind,+--pow)
++++++++{
++++++++++++var+cVal+=+Convert.ToInt32(colLabel[ind])+-+64;+//col+A+is+index+1
++++++++++++colIndex+%2B=+cVal+*+((int)Math.Pow(26,+pow));
++++++++}
++++++++return+colIndex;
++++}
}|code-block|syntax|javascript|3953766|像这样使用它。|3953767|30.ColumnLabel();+//+"AD"
"AD".ColumnIndex();+//+30|3953768|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I wanted to throw in my static class I use, for interoping between col index and col Label. I use a modified accepted answer for my ColumnLabel Method

<pre><code>public static class Extensions
{
 public static string ColumnLabel(this int col)
 {
 var dividend = col;
 var columnLabel = string.Empty;
 int modulo;

 while (dividend &gt; 0)
 {
 modulo = (dividend - 1) % 26;
 columnLabel = Convert.ToChar(65 + modulo).ToString() + columnLabel;
 dividend = (int)((dividend - modulo) / 26);
 } 

 return columnLabel;
 }
 public static int ColumnIndex(this string colLabel)
 {
 // "AD" (1 * 26^1) + (4 * 26^0) ...
 var colIndex = 0;
 for(int ind = 0, pow = colLabel.Count()-1; ind &lt; colLabel.Count(); ++ind, --pow)
 {
 var cVal = Convert.ToInt32(colLabel[ind]) - 64; //col A is index 1
 colIndex += cVal * ((int)Math.Pow(26, pow));
 }
 return colIndex;
 }
}
</code></pre>

Use this like...

<pre><code>30.ColumnLabel(); // "AD"
"AD".ColumnIndex(); // 30
</code></pre>

blocks|key|806208|text|private+String+getColumn(int+c)+{
++++String+s+=+"";
++++do+{
++++++++s+=+(char)('A'+%2B+(c+%25+26))+%2B+s;
++++++++c+/=+26;
++++}+while+(c--+>+0);
++++return+s;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|806209|它不是精确的基数26，系统中没有0。如果有，“Z”后面应该是“BA”，而不是“AA”。|unstyled|806210|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>private String getColumn(int c) {
 String s = "";
 do {
 s = (char)('A' + (c % 26)) + s;
 c /= 26;
 } while (c-- &gt; 0);
 return s;
}
</code></pre>

Its not exactly base 26, there is no 0 in the system. If there was, 'Z' would be followed by 'BA' not by 'AA'.

blocks|key|806436|text|如果你只是想要一个没有代码的单元格公式，这里有一个公式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|806437|IF(COLUMN()>=26,CHAR(ROUND(COLUMN()/26,1)%2B64)&CHAR(MOD(COLUMN(),26)%2B64),CHAR(COLUMN()%2B64))|code-block|syntax|javascript|806438|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

if you just want it for a cell formula without code, here's a formula for it:

<pre><code>IF(COLUMN()&gt;=26,CHAR(ROUND(COLUMN()/26,1)+64)&amp;CHAR(MOD(COLUMN(),26)+64),CHAR(COLUMN()+64))
</code></pre>

blocks|key|3107620|text|在Delphi+(Pascal)中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3107621|function+GetExcelColumnName(columnNumber:+integer):+string;
var
++dividend,+modulo:+integer;
begin
++Result+:=+'';
++dividend+:=+columnNumber;
++while+dividend+>+0+do+begin
++++modulo+:=+(dividend+-+1)+mod+26;
++++Result+:=+Chr(65+%2B+modulo)+%2B+Result;
++++dividend+:=+(dividend+-+modulo)+div+26;
++end;
end;|code-block|syntax|javascript|3107622|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In Delphi (Pascal):

<pre><code>function GetExcelColumnName(columnNumber: integer): string;
var
 dividend, modulo: integer;
begin
 Result := '';
 dividend := columnNumber;
 while dividend &gt; 0 do begin
 modulo := (dividend - 1) mod 26;
 Result := Chr(65 + modulo) + Result;
 dividend := (dividend - modulo) div 26;
 end;
end;
</code></pre>

blocks|key|5363109|text|有点晚了，但下面是我使用的代码(在C#中)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5363110|private+static+readonly+string+_Alphabet+=+"ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public+static+int+ColumnNameParse(string+value)
{
++++//+assumes+value.Length+is+[1,3]
++++//+assumes+value+is+uppercase
++++var+digits+=+value.PadLeft(3).Select(x+=>+_Alphabet.IndexOf(x));
++++return+digits.Aggregate(0,+(current,+index)+=>+(current+*+26)+%2B+(index+%2B+1));
}|code-block|syntax|javascript|5363111|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A little late to the game, but here's the code I use (in C#):

<pre><code>private static readonly string _Alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public static int ColumnNameParse(string value)
{
 // assumes value.Length is [1,3]
 // assumes value is uppercase
 var digits = value.PadLeft(3).Select(x =&gt; _Alphabet.IndexOf(x));
 return digits.Aggregate(0, (current, index) =&gt; (current * 26) + (index + 1));
}
</code></pre>

blocks|key|3107826|text|在perl语言中，输入1+(A)、27+(AA)等。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3107827|sub+excel_colname+{
++my+($idx)+=+@_;+++++++#+one-based+column+number
++--$idx;+++++++++++++++#+zero-based+column+index
++my+$name+=+"";
++while+($idx+>=+0)+{
++++$name+.=+chr(ord("A")+%2B+($idx+%25+26));
++++$idx+++=+int($idx+/+26)+-+1;
++}
++return+scalar+reverse+$name;
}|code-block|syntax|javascript|3107828|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In perl, for an input of 1 (A), 27 (AA), etc.

<pre><code>sub excel_colname {
 my ($idx) = @_; # one-based column number
 --$idx; # zero-based column index
 my $name = "";
 while ($idx &gt;= 0) {
 $name .= chr(ord("A") + ($idx % 26));
 $idx = int($idx / 26) - 1;
 }
 return scalar reverse $name;
}
</code></pre>

blocks|key|810911|text|在看了这里提供的所有版本之后，我决定自己做一个，使用递归。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810912|以下是我的vb.net版本：|810913|Function+CL(ByVal+x+As+Integer)+As+String
++++If+x+>=+1+And+x+<=+26+Then
++++++++CL+=+Chr(x+%2B+64)
++++Else
++++++++CL+=+CL((x+-+x+Mod+26)+/+26)+&+Chr((x+Mod+26)+%2B+1+%2B+64)
++++End+If
End+Function|code-block|syntax|javascript|810914|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

After looking at all the supplied Versions here, I decided to do one myself, using recursion.
Here is my vb.net Version:
<pre><code>Function CL(ByVal x As Integer) As String
 If x &gt;= 1 And x &lt;= 26 Then
 CL = Chr(x + 64)
 Else
 CL = CL((x - x Mod 26) / 26) &amp; Chr((x Mod 26) + 1 + 64)
 End If
End Function
</code></pre>

blocks|key|5363890|text|虽然我玩得很晚，但Graham's+answer远不是最优的。特别是，您不必使用modulo、调用ToString()和应用(int)强制转换。考虑到在C#世界中的大多数情况下，您将从0开始编号，以下是我的修订：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5363891|public+static+string+GetColumnName(int+index)+//+zero-based
{
++++const+byte+BASE+=+'Z'+-+'A'+%2B+1;
++++string+name+=+String.Empty;

++++do
++++{
++++++++name+=+Convert.ToChar('A'+%2B+index+%25+BASE)+%2B+name;
++++++++index+=+index+/+BASE+-+1;
++++}
++++while+(index+>=+0);

++++return+name;
}|code-block|syntax|javascript|5363892|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/182924/1997693^0|14|6|1D|A|1Q|5|9|F|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Though I am late to the game, <a href="https://stackoverflow.com/a/182924/1997693">Graham's answer</a> is far from being optimal. Particularly, you don't have to use the <code>modulo</code>, call <code>ToString()</code> and apply <code>(int)</code> cast. Considering that in most cases in C# world you would start numbering from 0, here is my revision:

<pre class="lang-cs prettyprint-override"><code>public static string GetColumnName(int index) // zero-based
{
 const byte BASE = 'Z' - 'A' + 1;
 string name = String.Empty;

 do
 {
 name = Convert.ToChar('A' + index % BASE) + name;
 index = index / BASE - 1;
 }
 while (index &gt;= 0);

 return name;
}
</code></pre>

blocks|key|3954634|text|已经有超过30种解决方案了，但这里是我的一行C#解决方案……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3954635|public+string+IntToExcelColumn(int+i)
{
++++return+((i<16926?+""+:+((char)((((i/26)-1)%2526)%2B65)).ToString())+%2B+(i<2730?+""+:+((char)((((i/26)-1)%2526)%2B65)).ToString())+%2B+(i<26?+""+:+((char)((((i/26)-1)%2526)%2B65)).ToString())+%2B+((char)((i%2526)%2B65)));
}|code-block|syntax|javascript|3954636|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

More than 30 solutions already, but here's my one-line C# solution... 

<pre><code>public string IntToExcelColumn(int i)
{
 return ((i&lt;16926? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i&lt;2730? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i&lt;26? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + ((char)((i%26)+65)));
}
</code></pre>

blocks|key|3107182|text|优化原始解决方案(在C#中)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3107183|public+static+class+ExcelHelper
{
++++private+static+Dictionary<UInt16,+String>+l_DictionaryOfColumns;

++++public+static+ExcelHelper()+{
++++++++l_DictionaryOfColumns+=+new+Dictionary<ushort,+string>(256);
++++}

++++public+static+String+GetExcelColumnName(UInt16+l_Column)
++++{
++++++++UInt16+l_ColumnCopy+=+l_Column;
++++++++String+l_Chars+=+"0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
++++++++String+l_rVal+=+"";
++++++++UInt16+l_Char;


++++++++if+(l_DictionaryOfColumns.ContainsKey(l_Column)+==+true)
++++++++{
++++++++++++l_rVal+=+l_DictionaryOfColumns[l_Column];
++++++++}
++++++++else
++++++++{
++++++++++++while+(l_ColumnCopy+>+26)
++++++++++++{
++++++++++++++++l_Char+=+l_ColumnCopy+%25+26;
++++++++++++++++if+(l_Char+==+0)
++++++++++++++++++++l_Char+=+26;

++++++++++++++++l_ColumnCopy+=+(l_ColumnCopy+-+l_Char)+/+26;
++++++++++++++++l_rVal+=+l_Chars[l_Char]+%2B+l_rVal;
++++++++++++}
++++++++++++if+(l_ColumnCopy+!=+0)
++++++++++++++++l_rVal+=+l_Chars[l_ColumnCopy]+%2B+l_rVal;

++++++++++++l_DictionaryOfColumns.ContainsKey(l_Column)+=+l_rVal;
++++++++}

++++++++return+l_rVal;
++++}
}|code-block|syntax|javascript|3107184|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Refining the original solution (in C#):

<pre><code>public static class ExcelHelper
{
 private static Dictionary&lt;UInt16, String&gt; l_DictionaryOfColumns;

 public static ExcelHelper() {
 l_DictionaryOfColumns = new Dictionary&lt;ushort, string&gt;(256);
 }

 public static String GetExcelColumnName(UInt16 l_Column)
 {
 UInt16 l_ColumnCopy = l_Column;
 String l_Chars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
 String l_rVal = "";
 UInt16 l_Char;


 if (l_DictionaryOfColumns.ContainsKey(l_Column) == true)
 {
 l_rVal = l_DictionaryOfColumns[l_Column];
 }
 else
 {
 while (l_ColumnCopy &gt; 26)
 {
 l_Char = l_ColumnCopy % 26;
 if (l_Char == 0)
 l_Char = 26;

 l_ColumnCopy = (l_ColumnCopy - l_Char) / 26;
 l_rVal = l_Chars[l_Char] + l_rVal;
 }
 if (l_ColumnCopy != 0)
 l_rVal = l_Chars[l_ColumnCopy] + l_rVal;

 l_DictionaryOfColumns.ContainsKey(l_Column) = l_rVal;
 }

 return l_rVal;
 }
}
</code></pre>

blocks|key|3953018|text|下面是一个Actionscript版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3953019|private+var+columnNumbers:Array+=+['A',+'B',+'C',+'D',+'E',+'F'+,+'G',+'H',+'I',+'J',+'K'+,'L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];

++++private+function+getExcelColumnName(columnNumber:int)+:+String{
++++++++var+dividend:int+=+columnNumber;
++++++++var+columnName:String+=+"";
++++++++var+modulo:int;

++++++++while+(dividend+>+0)
++++++++{
++++++++++++modulo+=+(dividend+-+1)+%25+26;
++++++++++++columnName+=+columnNumbers[modulo]+%2B+columnName;
++++++++++++dividend+=+int((dividend+-+modulo)+/+26);
++++++++}+

++++++++return+columnName;
++++}|code-block|syntax|javascript|3953020|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is an Actionscript version:

<pre><code>private var columnNumbers:Array = ['A', 'B', 'C', 'D', 'E', 'F' , 'G', 'H', 'I', 'J', 'K' ,'L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];

 private function getExcelColumnName(columnNumber:int) : String{
 var dividend:int = columnNumber;
 var columnName:String = "";
 var modulo:int;

 while (dividend &gt; 0)
 {
 modulo = (dividend - 1) % 26;
 columnName = columnNumbers[modulo] + columnName;
 dividend = int((dividend - modulo) / 26);
 } 

 return columnName;
 }
</code></pre>

blocks|key|5363399|text|JavaScript解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|5363400|/**
+*+Calculate+the+column+letter+abbreviation+from+a+1+based+index
+*+@param+{Number}+value
+*+@returns+{string}
+*/
getColumnFromIndex+=+function+(value)+{
++++var+base+=+'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split('');
++++var+remainder,+result+=+"";
++++do+{
++++++++remainder+=+value+%25+26;
++++++++result+=+base[(remainder+%7C%7C+26)+-+1]+%2B+result;
++++++++value+=+Math.floor(value+/+26);
++++}+while+(value+>+0);
++++return+result;
};|code-block|syntax|javascript|5363401|entityMap^0|0|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

JavaScript Solution

<pre><code>/**
 * Calculate the column letter abbreviation from a 1 based index
 * @param {Number} value
 * @returns {string}
 */
getColumnFromIndex = function (value) {
 var base = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split('');
 var remainder, result = "";
 do {
 remainder = value % 26;
 result = base[(remainder || 26) - 1] + result;
 value = Math.floor(value / 26);
 } while (value &gt; 0);
 return result;
};
</code></pre>

blocks|key|806801|text|另一种VBA方式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|806802|Public+Function+GetColumnName(TargetCell+As+Range)+As+String
++++GetColumnName+=+Split(CStr(TargetCell.Cells(1,+1).Address),+"$")(1)
End+Function|code-block|syntax|javascript|806803|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another VBA way

<pre><code>Public Function GetColumnName(TargetCell As Range) As String
 GetColumnName = Split(CStr(TargetCell.Cells(1, 1).Address), "$")(1)
End Function
</code></pre>

blocks|key|3108282|text|这些我的代码可以将特定的数字(索引从1开始)转换为Excel列。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3108283|++++public+static+string+NumberToExcelColumn(uint+number)
++++{
++++++++uint+originalNumber+=+number;

++++++++uint+numChars+=+1;
++++++++while+(Math.Pow(26,+numChars)+<+number)
++++++++{
++++++++++++numChars%2B%2B;

++++++++++++if+(Math.Pow(26,+numChars)+%2B+26+>=+number)
++++++++++++{
++++++++++++++++break;
++++++++++++}+++++++++++++++
++++++++}

++++++++string+toRet+=+"";
++++++++uint+lastValue+=+0;

++++++++do
++++++++{
++++++++++++number+-=+lastValue;

++++++++++++double+powerVal+=+Math.Pow(26,+numChars+-+1);
++++++++++++byte+thisCharIdx+=+(byte)Math.Truncate((columnNumber+-+1)+/+powerVal);
++++++++++++lastValue+=+(int)powerVal+*+thisCharIdx;

++++++++++++if+(numChars+-+2+>=+0)
++++++++++++{
++++++++++++++++double+powerVal_next+=+Math.Pow(26,+numChars+-+2);
++++++++++++++++byte+thisCharIdx_next+=+(byte)Math.Truncate((columnNumber+-+lastValue+-+1)+/+powerVal_next);
++++++++++++++++int+lastValue_next+=+(int)Math.Pow(26,+numChars+-+2)+*+thisCharIdx_next;

++++++++++++++++if+(thisCharIdx_next+==+0+&&+lastValue_next+==+0+&&+powerVal_next+==+26)
++++++++++++++++{
++++++++++++++++++++thisCharIdx--;
++++++++++++++++++++lastValue+=+(int)powerVal+*+thisCharIdx;
++++++++++++++++}
++++++++++++}

++++++++++++toRet+%2B=+(char)((byte)'A'+%2B+thisCharIdx+%2B+((numChars+>+1)+?+-1+:+0));

++++++++++++numChars--;
++++++++}+while+(numChars+>+0);

++++++++return+toRet;
++++}|code-block|syntax|javascript|3108284|我的单元测试：|3108285|++++[TestMethod]
++++public+void+Test()
++++{
++++++++Assert.AreEqual("A",+NumberToExcelColumn(1));
++++++++Assert.AreEqual("Z",+NumberToExcelColumn(26));
++++++++Assert.AreEqual("AA",+NumberToExcelColumn(27));
++++++++Assert.AreEqual("AO",+NumberToExcelColumn(41));
++++++++Assert.AreEqual("AZ",+NumberToExcelColumn(52));
++++++++Assert.AreEqual("BA",+NumberToExcelColumn(53));
++++++++Assert.AreEqual("ZZ",+NumberToExcelColumn(702));
++++++++Assert.AreEqual("AAA",+NumberToExcelColumn(703));
++++++++Assert.AreEqual("ABC",+NumberToExcelColumn(731));
++++++++Assert.AreEqual("ACQ",+NumberToExcelColumn(771));
++++++++Assert.AreEqual("AYZ",+NumberToExcelColumn(1352));
++++++++Assert.AreEqual("AZA",+NumberToExcelColumn(1353));
++++++++Assert.AreEqual("AZB",+NumberToExcelColumn(1354));
++++++++Assert.AreEqual("BAA",+NumberToExcelColumn(1379));
++++++++Assert.AreEqual("CNU",+NumberToExcelColumn(2413));
++++++++Assert.AreEqual("GCM",+NumberToExcelColumn(4823));
++++++++Assert.AreEqual("MSR",+NumberToExcelColumn(9300));
++++++++Assert.AreEqual("OMB",+NumberToExcelColumn(10480));
++++++++Assert.AreEqual("ULV",+NumberToExcelColumn(14530));
++++++++Assert.AreEqual("XFD",+NumberToExcelColumn(16384));
++++}|3108286|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

These my codes to convert specific number (index start from 1) to Excel Column.

<pre><code> public static string NumberToExcelColumn(uint number)
 {
 uint originalNumber = number;

 uint numChars = 1;
 while (Math.Pow(26, numChars) &lt; number)
 {
 numChars++;

 if (Math.Pow(26, numChars) + 26 &gt;= number)
 {
 break;
 } 
 }

 string toRet = "";
 uint lastValue = 0;

 do
 {
 number -= lastValue;

 double powerVal = Math.Pow(26, numChars - 1);
 byte thisCharIdx = (byte)Math.Truncate((columnNumber - 1) / powerVal);
 lastValue = (int)powerVal * thisCharIdx;

 if (numChars - 2 &gt;= 0)
 {
 double powerVal_next = Math.Pow(26, numChars - 2);
 byte thisCharIdx_next = (byte)Math.Truncate((columnNumber - lastValue - 1) / powerVal_next);
 int lastValue_next = (int)Math.Pow(26, numChars - 2) * thisCharIdx_next;

 if (thisCharIdx_next == 0 &amp;&amp; lastValue_next == 0 &amp;&amp; powerVal_next == 26)
 {
 thisCharIdx--;
 lastValue = (int)powerVal * thisCharIdx;
 }
 }

 toRet += (char)((byte)'A' + thisCharIdx + ((numChars &gt; 1) ? -1 : 0));

 numChars--;
 } while (numChars &gt; 0);

 return toRet;
 }
</code></pre>

My Unit Test:

<pre><code> [TestMethod]
 public void Test()
 {
 Assert.AreEqual("A", NumberToExcelColumn(1));
 Assert.AreEqual("Z", NumberToExcelColumn(26));
 Assert.AreEqual("AA", NumberToExcelColumn(27));
 Assert.AreEqual("AO", NumberToExcelColumn(41));
 Assert.AreEqual("AZ", NumberToExcelColumn(52));
 Assert.AreEqual("BA", NumberToExcelColumn(53));
 Assert.AreEqual("ZZ", NumberToExcelColumn(702));
 Assert.AreEqual("AAA", NumberToExcelColumn(703));
 Assert.AreEqual("ABC", NumberToExcelColumn(731));
 Assert.AreEqual("ACQ", NumberToExcelColumn(771));
 Assert.AreEqual("AYZ", NumberToExcelColumn(1352));
 Assert.AreEqual("AZA", NumberToExcelColumn(1353));
 Assert.AreEqual("AZB", NumberToExcelColumn(1354));
 Assert.AreEqual("BAA", NumberToExcelColumn(1379));
 Assert.AreEqual("CNU", NumberToExcelColumn(2413));
 Assert.AreEqual("GCM", NumberToExcelColumn(4823));
 Assert.AreEqual("MSR", NumberToExcelColumn(9300));
 Assert.AreEqual("OMB", NumberToExcelColumn(10480));
 Assert.AreEqual("ULV", NumberToExcelColumn(14530));
 Assert.AreEqual("XFD", NumberToExcelColumn(16384));
 }
</code></pre>

blocks|key|3108737|text|这是所有其他人以及谷歌重定向到的问题，所以我在这里张贴这篇文章。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3108738|这些答案中的许多都是正确的，但对于简单的情况来说太麻烦了，比如当您的列不超过26列时。如果你有任何疑问，你是否会进入双字符列，然后忽略这个答案，但是如果你确定你不会，那么你可以在C#中像这样简单地完成：|3108739|public+static+char+ColIndexToLetter(short+index)
{
++++if+(index+<+0+%7C%7C+index+>+25)+throw+new+ArgumentException("Index+must+be+between+0+and+25.");
++++return+(char)('A'+%2B+index);
}|code-block|syntax|javascript|3108740|见鬼，如果你对你正在传递的东西有信心，你甚至可以删除验证，并使用下面的内联：|3108741|(char)('A'+%2B+index)|3108742|这在许多语言中都非常相似，因此您可以根据需要对其进行调整。|3108743|同样，只有当你100%25确定你的不会超过26列时，才会使用它。|offset|length|style|BOLD|3108744|entityMap^0|0|0|0|0|0|0|O|6|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

This is the question all others as well as Google redirect to so I'm posting this here.

Many of these answers are correct but too cumbersome for simple situations such as when you don't have over 26 columns. If you have any doubt whether you might go into double character columns then ignore this answer, but if you're sure you won't, then you could do it as simple as this in C#:

<pre><code>public static char ColIndexToLetter(short index)
{
 if (index &lt; 0 || index &gt; 25) throw new ArgumentException("Index must be between 0 and 25.");
 return (char)('A' + index);
}
</code></pre>

Heck, if you're confident about what you're passing in you could even remove the validation and use this inline:

<pre><code>(char)('A' + index)
</code></pre>

This will be very similar in many languages so you can adapt it as needed.

Again, only use this if you're 100% sure you won't have more than 26 columns.

blocks|key|3954404|text|对不起，这是Python而不是C#，但至少结果是正确的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3954405|def+excel_column_number_to_name(column_number):
++++output+=+""
++++index+=+column_number-1
++++while+index+>=+0:
++++++++character+=+chr((index%2526)%2Bord('A'))
++++++++output+=+output+%2B+character
++++++++index+=+index/26+-+1

++++return+output[::-1]


for+i+in+xrange(1,+1024):
++++print+"%254d+:+%25s"+%25+(i,+excel_column_number_to_name(i))|code-block|syntax|javascript|3954406|通过以下测试用例：|3954407|3954408|列号:+494286+=>+|unordered-list-item|3954409|列号:+27+=>+AA|3954410|列号:+52+=>+AZ|3954411|3954412|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|X|8|@]|9|@]|A|$]]|$1|M|3|N|5|L|7|Y|8|@]|9|@]|A|$]]|$1|O|3|P|5|L|7|Z|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|S|$]]

Sorry, this is Python instead of C#, but at least the results are correct:

<pre><code>def excel_column_number_to_name(column_number):
 output = ""
 index = column_number-1
 while index &gt;= 0:
 character = chr((index%26)+ord('A'))
 output = output + character
 index = index/26 - 1

 return output[::-1]


for i in xrange(1, 1024):
 print "%4d : %s" % (i, excel_column_number_to_name(i))
</code></pre>

Passed these test cases:

<ul>
<li>Column Number: 494286 => ABCDZ </li>
<li>Column Number: 27 => AA</li>
<li>Column Number: 52 => AZ</li>
</ul>

blocks|key|5364406|text|下面是Graham在Powershell中的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5364407|function+ConvertTo-ExcelColumnID+{
param+(
++++[parameter(Position+=+0,
++++++++HelpMessage+=+"A+1-based+index+to+convert+to+an+excel+column+ID.+e.g.+2+=>+'B',+29+=>+'AC'",
++++++++Mandatory+=+$true)]
++++[int]$index
);

[string]$result+=+'';
if+($index+-le+0+)+{
++++return+$result;
}

while+($index+-gt+0)+{
++++[int]$modulo+=+($index+-+1)+%25+26;
++++$character+=+[char]($modulo+%2B+[int][char]'A');
++++$result+=+$character+%2B+$result;
++++[int]$index+=+($index+-+$modulo)+/+26;
}

return+$result;|code-block|syntax|javascript|5364408|}|5364409|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For what it is worth, here is Graham's code in Powershell:

<pre><code>function ConvertTo-ExcelColumnID {
param (
 [parameter(Position = 0,
 HelpMessage = "A 1-based index to convert to an excel column ID. e.g. 2 =&gt; 'B', 29 =&gt; 'AC'",
 Mandatory = $true)]
 [int]$index
);

[string]$result = '';
if ($index -le 0 ) {
 return $result;
}

while ($index -gt 0) {
 [int]$modulo = ($index - 1) % 26;
 $character = [char]($modulo + [int][char]'A');
 $result = $character + $result;
 [int]$index = ($index - $modulo) / 26;
}

return $result;
</code></pre>

}

blocks|key|3953402|text|我正试着用Java做同样的事情...我写了以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3953403|private+String+getExcelColumnName(int+columnNumber)+{

++++int+dividend+=+columnNumber;
++++String+columnName+=+"";
++++int+modulo;

++++while+(dividend+>+0)
++++{
++++++++modulo+=+(dividend+-+1)+%25+26;

++++++++char+val+=+Character.valueOf((char)(65+%2B+modulo));

++++++++columnName+%2B=+val;

++++++++dividend+=+(int)((dividend+-+modulo)+/+26);
++++}+

++++return+columnName;
}|code-block|syntax|javascript|3953404|现在，一旦我用columnNumber+=+29运行它，它给我的结果=+"CA“(而不是"AC")，有什么我遗漏的注释吗？我知道我可以通过StringBuilder来逆转它...但看着格雷厄姆的答案，我有点困惑...|3953405|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I'm trying to do the same thing in Java...
I've wrote following code:

<pre><code>private String getExcelColumnName(int columnNumber) {

 int dividend = columnNumber;
 String columnName = "";
 int modulo;

 while (dividend &gt; 0)
 {
 modulo = (dividend - 1) % 26;

 char val = Character.valueOf((char)(65 + modulo));

 columnName += val;

 dividend = (int)((dividend - modulo) / 26);
 } 

 return columnName;
}
</code></pre>

Now once I ran it with columnNumber = 29, it gives me the result = "CA" (instead of "AC")
any comments what I'm missing?
I know I can reverse it by StringBuilder.... But looking at the Graham's answer, I'm little confused....

blocks|key|3108044|text|下面是我用PHP实现的最新实现。这个是递归的。我是在找到这篇文章之前写的。我想看看其他人是否已经解决了这个问题……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3108045|public+function+GetColumn($intNumber,+$strCol+=+null)+{

++++if+($intNumber+>+0)+{
++++++++$intRem+=+($intNumber+-+1)+%25+26;
++++++++$strCol+=+$this->GetColumn(intval(($intNumber+-+$intRem)+/+26),+sprintf('%25s%25s',+chr(65+%2B+$intRem),+$strCol));
++++}

++++return+$strCol;
}|code-block|syntax|javascript|3108046|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's my super late implementation in PHP. This one's recursive. I wrote it just before I found this post. I wanted to see if others had solved this problem already...

<pre><code>public function GetColumn($intNumber, $strCol = null) {

 if ($intNumber &gt; 0) {
 $intRem = ($intNumber - 1) % 26;
 $strCol = $this-&gt;GetColumn(intval(($intNumber - $intRem) / 26), sprintf('%s%s', chr(65 + $intRem), $strCol));
 }

 return $strCol;
}
</code></pre>

blocks|key|5363693|text|简洁优雅的Ruby版：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5363694|def+col_name(col_idx)
++++name+=+""
++++while+col_idx>0
++++++++mod+++++=+(col_idx-1)%2526
++++++++name++++=+(65%2Bmod).chr+%2B+name
++++++++col_idx+=+((col_idx-mod)/26).to_i
++++end
++++name
end|code-block|syntax|javascript|5363695|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Coincise and elegant Ruby version:

<pre><code>def col_name(col_idx)
 name = ""
 while col_idx&gt;0
 mod = (col_idx-1)%26
 name = (65+mod).chr + name
 col_idx = ((col_idx-mod)/26).to_i
 end
 name
end
</code></pre>

blocks|key|811358|text|每种方式的F#版本|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811359|let+rec+getExcelColumnName+x++=+if+x<26+then+int+'A'%2Bx%7C>char%7C>string+else+(x/26-1%7C>c)%2B+c(x%2526)|code-block|syntax|javascript|811360|请原谅，我正在开发一个更好的https://stackoverflow.com/a/4500043/57883版本|offset|length|811361|​|811362|811363|和相反的方向：|811364|811365|//+return+values+start+at+0
let+getIndexFromExcelColumnName+(x:string)+=
++++let+a+=+int+'A'
++++let+fPow+len+i+=
++++++++Math.Pow(26.,+len+-+1+-+i+%7C>+float)
++++++++%7C>+int

++++let+getValue+len+i+c+=+
++++++++int+c+-+a+%2B+1+*+fPow+len+i
++++let+f+i+=+getValue+x.Length+i+x.[i]
++++[0+..+x.Length+-+1]
++++%7C>+Seq.map+f
++++%7C>+Seq.sum
++++%7C>+fun+x+->+x+-+1|811366|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/4500043/57883^0|0|0|E|15|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@$I|13|J|14|1|15]]|A|$]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|18|8|@]|9|@]|A|$]]|$1|P|3|L|5|6|7|19|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

F# version of each way

<pre><code>let rec getExcelColumnName x = if x&lt;26 then int 'A'+x|&gt;char|&gt;string else (x/26-1|&gt;c)+ c(x%26)
</code></pre>

pardon the minimizing, was working on a better version of <a href="https://stackoverflow.com/a/4500043/57883">https://stackoverflow.com/a/4500043/57883</a>

<hr />
and the opposite direction:

<pre><code>// return values start at 0
let getIndexFromExcelColumnName (x:string) =
 let a = int 'A'
 let fPow len i =
 Math.Pow(26., len - 1 - i |&gt; float)
 |&gt; int

 let getValue len i c = 
 int c - a + 1 * fPow len i
 let f i = getValue x.Length i x.[i]
 [0 .. x.Length - 1]
 |&gt; Seq.map f
 |&gt; Seq.sum
 |&gt; fun x -&gt; x - 1
</code></pre>

blocks|key|5364184|text|感谢你在这里的回答！帮助我想出了这些帮助函数，以便与我在Elixir/Phoenix中开发的Google+Sheets+API进行一些交互|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5364185|这是我想出来的(可能需要一些额外的验证和错误处理)|5364186|在Elixir中：|5364187|def+number_to_column(number)+do
++cond+do
++++(number+>+0+&&+number+<=+26)+->
++++++to_string([(number+%2B+64)])
++++(number+>+26)+->
++++++div_col+=+number_to_column(div(number+-+1,+26))
++++++remainder+=+rem(number,+26)
++++++rem_col+=+cond+do
++++++++(remainder+==+0)+->
++++++++++number_to_column(26)
++++++++true+->
++++++++++number_to_column(remainder)
++++++end
++++++div_col+<>+rem_col
++++true+->
++++++""
++end
end|code-block|syntax|javascript|5364188|和反函数：|5364189|def+column_to_number(column)+do
++column
++++%7C>+to_charlist
++++%7C>+Enum.reverse
++++%7C>+Enum.with_index
++++%7C>+Enum.reduce(0,+fn({char,+idx},+acc)+->
++++++((char+-+64)+*+:math.pow(26,idx))+%2B+acc
++++end)
++++%7C>+round
end|5364190|还有一些测试：|5364191|describe+"test+excel+functions"+do
++@excelTestData+[{"A",+1},+{"Z",26},+{"AA",+27},+{"AB",+28},+{"AZ",+52},{"BA",+53},+{"AAA",+703}]

++test+"column+to+number"+do
++++Enum.each(@excelTestData,+fn({input,+expected_result})+->
++++++actual_result+=+BulkOnboardingController.column_to_number(input)
++++++assert+actual_result+==+expected_result
++++end)
++end

++test+"number+to+column"+do
++++Enum.each(@excelTestData,+fn({expected_result,+input})+->
++++++actual_result+=+BulkOnboardingController.number_to_column(input)
++++++assert+actual_result+==+expected_result
++++end)
++end
end|5364192|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|H|7|11|8|@]|9|@]|A|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Thanks for the answers here!! helped me come up with these helper functions for some interaction with the Google Sheets API that i'm working on in Elixir/Phoenix

here's what i came up with (could probably use some extra validation and error handling)

In Elixir:

<pre><code>def number_to_column(number) do
 cond do
 (number &gt; 0 &amp;&amp; number &lt;= 26) -&gt;
 to_string([(number + 64)])
 (number &gt; 26) -&gt;
 div_col = number_to_column(div(number - 1, 26))
 remainder = rem(number, 26)
 rem_col = cond do
 (remainder == 0) -&gt;
 number_to_column(26)
 true -&gt;
 number_to_column(remainder)
 end
 div_col &lt;&gt; rem_col
 true -&gt;
 ""
 end
end
</code></pre>

And the inverse function: 

<pre><code>def column_to_number(column) do
 column
 |&gt; to_charlist
 |&gt; Enum.reverse
 |&gt; Enum.with_index
 |&gt; Enum.reduce(0, fn({char, idx}, acc) -&gt;
 ((char - 64) * :math.pow(26,idx)) + acc
 end)
 |&gt; round
end
</code></pre>

And some tests:

<pre><code>describe "test excel functions" do
 @excelTestData [{"A", 1}, {"Z",26}, {"AA", 27}, {"AB", 28}, {"AZ", 52},{"BA", 53}, {"AAA", 703}]

 test "column to number" do
 Enum.each(@excelTestData, fn({input, expected_result}) -&gt;
 actual_result = BulkOnboardingController.column_to_number(input)
 assert actual_result == expected_result
 end)
 end

 test "number to column" do
 Enum.each(@excelTestData, fn({expected_result, input}) -&gt;
 actual_result = BulkOnboardingController.number_to_column(input)
 assert actual_result == expected_result
 end)
 end
end
</code></pre>

blocks|key|3108949|text|这是根据Graham的代码编写的javascript版本|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3108950|function+(columnNumber)+{
++++var+dividend+=+columnNumber;
++++var+columnName+=+"";
++++var+modulo;

++++while+(dividend+>+0)+{
++++++++modulo+=+(dividend+-+1)+%25+26;
++++++++columnName+=+String.fromCharCode(65+%2B+modulo)+%2B+columnName;
++++++++dividend+=+parseInt((dividend+-+modulo)+/+26);
++++}

++++return+columnName;
};|code-block|syntax|javascript|3108951|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is a javascript version according to Graham's code

<pre><code>function (columnNumber) {
 var dividend = columnNumber;
 var columnName = "";
 var modulo;

 while (dividend &gt; 0) {
 modulo = (dividend - 1) % 26;
 columnName = String.fromCharCode(65 + modulo) + columnName;
 dividend = parseInt((dividend - modulo) / 26);
 }

 return columnName;
};
</code></pre>

blocks|key|5364499|text|以前的大多数答案都是正确的。这是另一种将列号转换为excel列的方法。如果我们认为这是一个基数转换，那么解决方案就相当简单。简单地将列数转换为基数26，因为这里只有26个字母。下面是你如何做到这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5364500|步骤：|offset|length|style|BOLD|5364501|从上一步开始，将列设置为quotient|5364502|subtract+one+from+|unordered-list-item|5364503|变量(从上一步开始)，因为我们需要以97为a结束ascii+table。|5364504|除以26，得到table)|5364505|quotient+%2B97的余数并转换为ascii+table(因为在ASCII中，97是"a“，因此|5364506|将成为新的商/+26+(因为我们可能需要超过26+column)|5364507|continue来执行此操作，直到商大于0，然后返回结果|5364508|5364509|5364510|5364511|下面是执行此操作的代码:)|5364512|def+convert_num_to_column(column_num):
++++result+=+""
++++quotient+=+column_num
++++remainder+=+0
++++while+(quotient+>0):
++++++++quotient+=+quotient+-1
++++++++remainder+=+quotient%2526
++++++++result+=+chr(int(remainder)%2B97)%2Bresult
++++++++quotient+=+int(quotient/26)
++++return+result

print("--",convert_num_to_column(1).upper())|code-block|syntax|javascript|5364513|entityMap|0|LINK|mutability|MUTABLE|url|http://www.asciitable.com/|1^0|0|0|3|0|0|0|O|B|0|0|0|J|B|1|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1G|8|@$D|1H|E|1I|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1J|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|1K|8|@]|9|@]|A|$]]|$1|M|3|N|5|L|7|1L|8|@]|9|@$D|1M|E|1N|1|1O]]|A|$]]|$1|O|3|P|5|L|7|1P|8|@]|9|@]|A|$]]|$1|Q|3|R|5|L|7|1Q|8|@]|9|@$D|1R|E|1S|1|1T]]|A|$]]|$1|S|3|T|5|6|7|1U|8|@]|9|@]|A|$]]|$1|U|3|V|5|L|7|1V|8|@]|9|@]|A|$]]|$1|W|3|-4|5|L|7|1W|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|11|3|12|5|13|7|20|8|@]|9|@]|A|$14|15]]|$1|16|3|-4|5|6|7|21|8|@]|9|@]|A|$]]]|17|$18|$5|19|1A|1B|A|$1C|1D]]|1E|$5|19|1A|1B|A|$1C|1D]]]]

Most of previous answers are correct. Here is one more way of converting column number to excel columns.
solution is rather simple if we think about this as a base conversion. Simply, convert the column number to base 26 since there is 26 letters only.
Here is how you can do this: 

steps:

<ul>
<li>set the column as a quotient</li>
<li>subtract one from quotient variable (from previous step) because we need to end up on <a href="http://www.asciitable.com/" rel="nofollow noreferrer">ascii table</a> with 97 being a.</li>
<li>divide by 26 and get the remainder.</li>
<li>add +97 to remainder and convert to char (since 97 is "a" in ASCII table)</li>
<li>quotient becomes the new quotient/ 26 (since we might go over 26 column)</li>
<li>continue to do this until quotient is greater than 0 and then return the result</li>
</ul>

here is the code that does this :)

<pre><code>def convert_num_to_column(column_num):
 result = ""
 quotient = column_num
 remainder = 0
 while (quotient &gt;0):
 quotient = quotient -1
 remainder = quotient%26
 result = chr(int(remainder)+97)+result
 quotient = int(quotient/26)
 return result

print("--",convert_num_to_column(1).upper())
</code></pre>

blocks|key|5364873|text|此代码段适用于A到ZZ列名|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5364874|string+columnName+=+columnNumber+>+26+?+Convert.ToChar(64+%2B+(columnNumber+/+26)).ToString()+%2B+Convert.ToChar(64+%2B+(columnNumber+%25+26))+:+Convert.ToChar(64+%2B+columnNumber).ToString();|code-block|syntax|javascript|5364875|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This snippet works for A to ZZ column Names
<pre><code>string columnName = columnNumber &gt; 26 ? Convert.ToChar(64 + (columnNumber / 26)).ToString() + Convert.ToChar(64 + (columnNumber % 26)) : Convert.ToChar(64 + columnNumber).ToString();
</code></pre>

blocks|key|5362450|text|我在VB.NET+2003中使用了这个，它工作得很好……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5362451|Private+Function+GetExcelColumnName(ByVal+aiColNumber+As+Integer)+As+String
++++Dim+BaseValue+As+Integer+=+Convert.ToInt32(("A").Chars(0))+-+1
++++Dim+lsReturn+As+String+=+String.Empty

++++If+(aiColNumber+>+26)+Then
++++++++lsReturn+=+GetExcelColumnName(Convert.ToInt32((Format(aiColNumber+/+26,+"0.0").Split("."))(0)))
++++End+If

++++GetExcelColumnName+=+lsReturn+%2B+Convert.ToChar(BaseValue+%2B+(aiColNumber+Mod+26))
End+Function|code-block|syntax|javascript|5362452|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I'm using this one in VB.NET 2003 and it works well...

<pre><code>Private Function GetExcelColumnName(ByVal aiColNumber As Integer) As String
 Dim BaseValue As Integer = Convert.ToInt32(("A").Chars(0)) - 1
 Dim lsReturn As String = String.Empty

 If (aiColNumber &gt; 26) Then
 lsReturn = GetExcelColumnName(Convert.ToInt32((Format(aiColNumber / 26, "0.0").Split("."))(0)))
 End If

 GetExcelColumnName = lsReturn + Convert.ToChar(BaseValue + (aiColNumber Mod 26))
End Function
</code></pre>

blocks|key|5362486|text|如果您希望以编程方式引用单元格，那么如果您使用工作表的Cells方法，您将获得更具可读性的代码。它采用行和列索引，而不是传统的单元格引用。它与Offset方法非常相似。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5362487|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

If you are wanting to reference the cell progmatically then you will get much more readable code if you use the Cells method of a sheet. It takes a row and column index instead of a traditonal cell reference. It is very similar to the Offset method.

blocks|key|5362561|text|在VB.Net+2005中使用此命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5362562|Private+Function+ColumnName(ByVal+ColumnIndex+As+Integer)+As+String

+++Dim+Name+As+String+=+""

+++Name+=+(New+Microsoft.Office.Interop.Owc11.Spreadsheet).Columns.Item(ColumnIndex).Address(False,+False,+Microsoft.Office.Interop.Owc11.XlReferenceStyle.xlA1)
+++Name+=+Split(Name,+":")(0)

+++Return+Name

End+Function|code-block|syntax|javascript|5362563|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using this in VB.Net 2005 :

<pre><code>Private Function ColumnName(ByVal ColumnIndex As Integer) As String

 Dim Name As String = ""

 Name = (New Microsoft.Office.Interop.Owc11.Spreadsheet).Columns.Item(ColumnIndex).Address(False, False, Microsoft.Office.Interop.Owc11.XlReferenceStyle.xlA1)
 Name = Split(Name, ":")(0)

 Return Name

End Function
</code></pre>

blocks|key|810721|text|另一种解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|810722|private+void+Foo()
{
+++l_ExcelApp+=+new+Excel.ApplicationClass();
+++l_ExcelApp.ReferenceStyle+=+Excel.XlReferenceStyle.xlR1C1;
+++//+...+now+reference+by+R[row]C[column],+Ex.+A1+<==>+R1C1,+C6+<==>+R3C6,+...
}|code-block|syntax|javascript|810723|更多信息请点击此处-+Cell+referencing+in+Excel+for+everyone!+by+Dr+Nitin+Paranjape|offset|length|810724|entityMap|0|LINK|mutability|MUTABLE|url|http://www.expresscomputeronline.com/20021216/techspace1.shtml^0|0|0|B|1P|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Another solution:

<pre><code>private void Foo()
{
 l_ExcelApp = new Excel.ApplicationClass();
 l_ExcelApp.ReferenceStyle = Excel.XlReferenceStyle.xlR1C1;
 // ... now reference by R[row]C[column], Ex. A1 &lt;==&gt; R1C1, C6 &lt;==&gt; R3C6, ...
}
</code></pre>

see more here - <a href="http://www.expresscomputeronline.com/20021216/techspace1.shtml" rel="nofollow noreferrer">Cell referencing in Excel for everyone! by Dr Nitin Paranjape</a>

blocks|key|806357|text|public+static+string+ConvertToAlphaColumnReferenceFromInteger(int+columnReference)
++++{
++++++++int+baseValue+=+((int)('A'))+-+1+;
++++++++string+lsReturn+=+String.Empty;+

++++++++if+(columnReference+>+26)+
++++++++{
++++++++++++lsReturn+=+ConvertToAlphaColumnReferenceFromInteger(Convert.ToInt32(Convert.ToDouble(columnReference+/+26).ToString().Split('.')[0]));
++++++++}+

++++++++return+lsReturn+%2B+Convert.ToChar(baseValue+%2B+(columnReference+%25+26));++++++++++++
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|806358|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static string ConvertToAlphaColumnReferenceFromInteger(int columnReference)
 {
 int baseValue = ((int)('A')) - 1 ;
 string lsReturn = String.Empty; 

 if (columnReference &gt; 26) 
 {
 lsReturn = ConvertToAlphaColumnReferenceFromInteger(Convert.ToInt32(Convert.ToDouble(columnReference / 26).ToString().Split('.')[0]));
 } 

 return lsReturn + Convert.ToChar(baseValue + (columnReference % 26)); 
 }
</code></pre>

blocks|key|3953249|text|下面是我在Python中的实现方法。算法解释如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3953250|alph+=+('a',+'b',+'c',+'d',+'e',+'f',+'g',+'h',+'i',+'j',+'k',+'l',+'m',+'n',+'o',+'p',+'q',+'r',+'s',+'t',+'u',+'v',+'w',+'x',+'y',+'z')
def+labelrec(n,+res):
++++if+n<26:
++++++++return+alph[n]%2Bres
++++else:
++++++++rem+=+n%2526
++++++++res+=+alph[rem]%2Bres
++++++++n+=+n/26-1
++++++++return+labelrec(n,+res)|code-block|syntax|javascript|3953251|可以使用数字和空字符串调用函数labelrec，如下所示：|3953252|print+labelrec(16383,+'')|3953253|这就是它的工作原理:如果十进制数的写入方式与Excel表列相同，则数字0-9将正常写入，但10将变为'00‘，然后20将变为'10’，依此类推。映射几个数字：|3953254|0-0|3953255|9-9|3953256|10+-+00|3953257|20+-+10|3953258|100+-+90|3953259|110+-+000|3953260|1110+-+0000|3953261|因此，模式是明确的。从单元的位置开始，如果一个数字小于10，它的表示与数字本身相同，否则您需要通过减去1和递归来调整剩余的数字。当数字小于10时，您可以停止。|3953262|在上面的解决方案中，相同的逻辑应用于基数为26的数字。|3953263|附注:如果希望数字从1开始，请在输入数字减去1后对其调用相同的函数。|3953264|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1D|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1F|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1J|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|1K|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|1L|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|1M|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|17|$]]

Here is how I would do it in Python. The algorithm is explained below:

<pre><code>alph = ('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z')
def labelrec(n, res):
 if n&lt;26:
 return alph[n]+res
 else:
 rem = n%26
 res = alph[rem]+res
 n = n/26-1
 return labelrec(n, res)
</code></pre>

The function labelrec can be called with the number and an empty string like:

<pre><code>print labelrec(16383, '')
</code></pre>

Here is why it works:
If decimal numbers were written the same way as Excel sheet columns, number 0-9 would be written normally, but 10 would become '00' and then 20 would become '10' and so on. Mapping few numbers:

0 - 0

9 - 9

10 - 00

20 - 10

100 - 90

110 - 000

1110 - 0000

So, the pattern is clear. Starting at the unit's place, if a number is less than 10, it's representation is same as the number itself, else you need to adjust the remaining number by subtracting it by 1 and recurse. You can stop when the number is less than 10.

The same logic is applied for numbers of base 26 in above solution.

P.S. If you want the numbers to begin from 1, call the same function on input number after decreasing it by 1.

blocks|key|811101|text|(我意识到这个问题与C#有关，但是，如果任何读者需要用Java做同样的事情，那么下面的内容可能会很有用)|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|811102|事实证明，使用Jakarta+POI中的"CellReference“类可以很容易地做到这一点。此外，转换可以通过两种方式完成。|811103|//+Convert+row+and+column+numbers+(0-based)+to+an+Excel+cell+reference
CellReference+numbers+=+new+CellReference(3,+28);
System.out.println(numbers.formatAsString());

//+Convert+an+Excel+cell+reference+back+into+digits
CellReference+reference+=+new+CellReference("AC4");
System.out.println(reference.getRow()+%2B+",+"+%2B+reference.getCol());|code-block|syntax|javascript|811104|entityMap|0|LINK|mutability|MUTABLE|url|http://poi.apache.org/^0|R|4|0|7|B|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|H|3|I|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

(I realise the question relates to C# however, if anyone reading needs to do the same thing with Java then the following may be useful)

It turns out that this can easily be done using the the "CellReference" class in <a href="http://poi.apache.org/" rel="nofollow">Jakarta POI</a>. Also, the conversion can be done both ways.

<pre><code>// Convert row and column numbers (0-based) to an Excel cell reference
CellReference numbers = new CellReference(3, 28);
System.out.println(numbers.formatAsString());

// Convert an Excel cell reference back into digits
CellReference reference = new CellReference("AC4");
System.out.println(reference.getRow() + ", " + reference.getCol());
</code></pre>

blocks|key|811227|text|我今天必须完成这项工作，我的实现使用了递归：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811228|private+static+string+GetColumnLetter(string+colNumber)
{
++++if+(string.IsNullOrEmpty(colNumber))
++++{
++++++++throw+new+ArgumentNullException(colNumber);
++++}

++++string+colName+=+String.Empty;

++++try
++++{
++++++++var+colNum+=+Convert.ToInt32(colNumber);
++++++++var+mod+=+colNum+%25+26;
++++++++var+div+=+Math.Floor((double)(colNum)/26);
++++++++colName+=+((div+>+0)+?+GetColumnLetter((div+-+1).ToString())+:+String.Empty)+%2B+Convert.ToChar(mod+%2B+65);
++++}
++++finally
++++{
++++++++colName+=+colName+==+String.Empty+?+"A"+:+colName;
++++}

++++return+colName;
}|code-block|syntax|javascript|811229|该方法将以字符串形式出现的数字和从"0“(A+=+0)开始的数字考虑在内。|811230|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I just had to do this work today, my implementation uses recursion:

<pre><code>private static string GetColumnLetter(string colNumber)
{
 if (string.IsNullOrEmpty(colNumber))
 {
 throw new ArgumentNullException(colNumber);
 }

 string colName = String.Empty;

 try
 {
 var colNum = Convert.ToInt32(colNumber);
 var mod = colNum % 26;
 var div = Math.Floor((double)(colNum)/26);
 colName = ((div &gt; 0) ? GetColumnLetter((div - 1).ToString()) : String.Empty) + Convert.ToChar(mod + 65);
 }
 finally
 {
 colName = colName == String.Empty ? "A" : colName;
 }

 return colName;
}
</code></pre>

This considers the number coming as string the the method and the numbers starting in "0" (A = 0)

blocks|key|3953946|text|Objective-C实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3953947|-(NSString*)getColumnName:(int)n+{
+++++NSString+*name+=+@"";
+++++while+(n>0)+{
+++++n--;
+++++char+c+=+(char)('A'+%2B+n%2526);
+++++name+=+[NSString+stringWithFormat:@"%25c%25@",c,name];
+++++n+=+n/26;
++}++++
+++++return+name;|code-block|syntax|javascript|3953948|}|3953949|SWIFT实施：|3953950|func+getColumnName(n:Int)->String{
+var+columnName+=+""
+var+index+=+n
+while+index>0+{
+++++index--
+++++let+char+=+Character(UnicodeScalar(65+%2B+index%2526))
+++++columnName+=+"\(char)\(columnName)"
+++++index+=+index+/+26
+}
+return+columnName|3953951|3953952|答案基于:https://stackoverflow.com/a/4532562/2231118|offset|length|3953953|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/4532562/2231118^0|0|0|0|0|0|0|5|17|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|11|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|12|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|M|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|15|8|@]|9|@$P|16|Q|17|1|18]]|A|$]]|$1|R|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y]]]]

Objective-C Implementation :

<pre><code>-(NSString*)getColumnName:(int)n {
 NSString *name = @"";
 while (n&gt;0) {
 n--;
 char c = (char)('A' + n%26);
 name = [NSString stringWithFormat:@"%c%@",c,name];
 n = n/26;
 } 
 return name;
</code></pre>

}

SWIFT Implementation:

<pre><code>func getColumnName(n:Int)-&gt;String{
 var columnName = ""
 var index = n
 while index&gt;0 {
 index--
 let char = Character(UnicodeScalar(65 + index%26))
 columnName = "\(char)\(columnName)"
 index = index / 26
 }
 return columnName
</code></pre>

}

The answer is based on :<a href="https://stackoverflow.com/a/4532562/2231118">https://stackoverflow.com/a/4532562/2231118</a>

blocks|key|3108441|text|NodeJS实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3108442|/**
*+getColumnFromIndex
*+Helper+that+returns+a+column+value+(A-XFD)+for+an+index+value+(integer).
*+The+column+follows+the+Common+Spreadsheet+Format+e.g.,+A,+AA,+AAA.
*+See+https://stackoverflow.com/questions/181596/how-to-convert-a-column-number-eg-127-into-an-excel-column-eg-aa/3444285#3444285
*+@param+numVal:+Integer
*+@return+String
*/
getColumnFromIndex:+function(numVal){
+++var+dividend+=+parseInt(numVal);
+++var+columnName+=+'';
+++var+modulo;
+++while+(dividend+>+0)+{
++++++modulo+=+(dividend+-+1)+%25+26;
++++++columnName+=+String.fromCharCode(65+%2B+modulo)+%2B+columnName;
++++++dividend+=+parseInt((dividend+-+modulo)+/+26);
+++}
+++return+columnName;
},|code-block|syntax|javascript|3108443|感谢Convert+excel+column+alphabet+(e.g.+AA)+to+number+(e.g.,+25)。反之：|offset|length|3108444|/**
*+getIndexFromColumn
*+Helper+that+returns+an+index+value+(integer)+for+a+column+value+(A-XFD).
*+The+column+follows+the+Common+Spreadsheet+Format+e.g.,+A,+AA,+AAA.
*+See+https://stackoverflow.com/questions/9905533/convert-excel-column-alphabet-e-g-aa-to-number-e-g-25
*+@param+strVal:+String
*+@return+Integer
*/
getIndexFromColumn:+function(val){
+++var+base+=+'ABCDEFGHIJKLMNOPQRSTUVWXYZ',+i,+j,+result+=+0;
+++for+(i+=+0,+j+=+val.length+-+1;+i+<+val.length;+i+%2B=+1,+j+-=+1)+{
++++++result+%2B=+Math.pow(base.length,+j)+*+(base.indexOf(val[i])+%2B+1);
+++}
+++return+result;
}|3108445|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/9905533/convert-excel-column-alphabet-e-g-aa-to-number-e-g-25^0|0|0|2|1O|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

NodeJS implementation:

<pre><code>/**
* getColumnFromIndex
* Helper that returns a column value (A-XFD) for an index value (integer).
* The column follows the Common Spreadsheet Format e.g., A, AA, AAA.
* See https://stackoverflow.com/questions/181596/how-to-convert-a-column-number-eg-127-into-an-excel-column-eg-aa/3444285#3444285
* @param numVal: Integer
* @return String
*/
getColumnFromIndex: function(numVal){
 var dividend = parseInt(numVal);
 var columnName = '';
 var modulo;
 while (dividend &gt; 0) {
 modulo = (dividend - 1) % 26;
 columnName = String.fromCharCode(65 + modulo) + columnName;
 dividend = parseInt((dividend - modulo) / 26);
 }
 return columnName;
},
</code></pre>

Thanks to <a href="https://stackoverflow.com/questions/9905533/convert-excel-column-alphabet-e-g-aa-to-number-e-g-25">Convert excel column alphabet (e.g. AA) to number (e.g., 25)</a>. And in reverse:

<pre><code>/**
* getIndexFromColumn
* Helper that returns an index value (integer) for a column value (A-XFD).
* The column follows the Common Spreadsheet Format e.g., A, AA, AAA.
* See https://stackoverflow.com/questions/9905533/convert-excel-column-alphabet-e-g-aa-to-number-e-g-25
* @param strVal: String
* @return Integer
*/
getIndexFromColumn: function(val){
 var base = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ', i, j, result = 0;
 for (i = 0, j = val.length - 1; i &lt; val.length; i += 1, j -= 1) {
 result += Math.pow(base.length, j) * (base.indexOf(val[i]) + 1);
 }
 return result;
}
</code></pre>

blocks|key|811290|text|我看到了另一个VBA答案--这可以在excel-vba中用一行自定义函数来完成：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|811291|Function+GetColLetter(ByVal+colID+As+Integer)+As+String
++++If+colID+>+Columns.Count+Then
++++++++Err.Raise+9,+,+"Column+index+out+of+bounds"
++++Else
++++++++GetColLetter+=+Split(Cells(1,+colID).Address,+"$")(1)
++++End+If
End+Function|code-block|syntax|javascript|811292|entityMap|0|LINK|mutability|MUTABLE|url|/questions/tagged/excel-vba^0|I|9|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Saw one other VBA answer - this can be done in <a href="/questions/tagged/excel-vba" class="post-tag" title="show questions tagged &#39;excel-vba&#39;" rel="tag">excel-vba</a> with a 1 line UDF:

<pre class="lang-vb prettyprint-override"><code>Function GetColLetter(ByVal colID As Integer) As String
 If colID &gt; Columns.Count Then
 Err.Raise 9, , &quot;Column index out of bounds&quot;
 Else
 GetColLetter = Split(Cells(1, colID).Address, &quot;$&quot;)(1)
 End If
End Function
</code></pre>

blocks|key|807271|text|Typescript|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|807272|function+lengthToExcelColumn(len:+number):+string+{

++++let+dividend:+number+=+len;
++++let+columnName:+string+=+'';
++++let+modulo:+number+=+0;

++++while+(dividend+>+0)+{
++++++++modulo+=+(dividend+-+1)+%25+26;
++++++++columnName+=+String.fromCharCode(65+%2B+modulo).toString()+%2B+columnName;
++++++++dividend+=+Math.floor((dividend+-+modulo)+/+26);
++++}
++++return+columnName;
}|code-block|syntax|javascript|807273|entityMap^0|0|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Typescript 

<pre><code>function lengthToExcelColumn(len: number): string {

 let dividend: number = len;
 let columnName: string = '';
 let modulo: number = 0;

 while (dividend &gt; 0) {
 modulo = (dividend - 1) % 26;
 columnName = String.fromCharCode(65 + modulo).toString() + columnName;
 dividend = Math.floor((dividend - modulo) / 26);
 }
 return columnName;
}
</code></pre>

blocks|key|807281|text|似乎有太多的答案比必要的复杂得多。下面是一个基于上述递归的通用Ruby答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|807282|这个答案的一个好处是它不限于英文字母表中的26个字符。你可以在COLUMNS常量中定义任何你喜欢的范围，它会做正确的事情。|offset|length|style|CODE|807283|++#+vim:+ft=ruby
++class+Numeric
++++COLUMNS+=+('A'..'Z').to_a

++++def+to_excel_column(n+=+self)
++++++n+<+1+?++''+:+begin
++++++++base+=+COLUMNS.size
++++++++to_excel_column((n+-+1)+/+base)+%2B+COLUMNS[(n+-+1)+%25+base]
++++++end
++++end
++end

++#+verify:
++(1..52).each+{+%7Ci%7C+printf+"%254d+=>+%254s\n",+i,+i.to_excel_column+}|code-block|syntax|javascript|807284|这将打印以下内容，例如：|807285|+++1+=>++++A
+++2+=>++++B
+++3+=>++++C
++....
++33+=>+++AG
++34+=>+++AH
++35+=>+++AI
++36+=>+++AJ
++37+=>+++AK
++38+=>+++AL
++39+=>+++AM
++40+=>+++AN
++41+=>+++AO
++42+=>+++AP
++43+=>+++AQ
++44+=>+++AR
++45+=>+++AS
++46+=>+++AT
++47+=>+++AU
++48+=>+++AV
++49+=>+++AW
++50+=>+++AX
++51+=>+++AY
++52+=>+++AZ|807286|entityMap^0|0|V|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Seems like so many answers are much more complex than necessary. Here is a generic Ruby answer based on the recursion described above:

One nice thing about this answer is that it's not limited to the 26 characters of English Alphabet. You can define any range you like in <code>COLUMNS</code> constant and it will do the right thing.

<pre><code> # vim: ft=ruby
 class Numeric
 COLUMNS = ('A'..'Z').to_a

 def to_excel_column(n = self)
 n &lt; 1 ? '' : begin
 base = COLUMNS.size
 to_excel_column((n - 1) / base) + COLUMNS[(n - 1) % base]
 end
 end
 end

 # verify:
 (1..52).each { |i| printf "%4d =&gt; %4s\n", i, i.to_excel_column }
</code></pre>

This prints the following, eg:

<pre><code> 1 =&gt; A
 2 =&gt; B
 3 =&gt; C
 ....
 33 =&gt; AG
 34 =&gt; AH
 35 =&gt; AI
 36 =&gt; AJ
 37 =&gt; AK
 38 =&gt; AL
 39 =&gt; AM
 40 =&gt; AN
 41 =&gt; AO
 42 =&gt; AP
 43 =&gt; AQ
 44 =&gt; AR
 45 =&gt; AS
 46 =&gt; AT
 47 =&gt; AU
 48 =&gt; AV
 49 =&gt; AW
 50 =&gt; AX
 51 =&gt; AY
 52 =&gt; AZ
</code></pre>

blocks|key|3109206|text|这是在编码测试中经常会问到的问题。它有一些约束:每个row=+702输出的最大列数应该具有行number%2Bcolumn名称，例如，703的答案是2A。(注意:我刚刚修改了另一个答案中的现有代码)下面是相同的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3109207|++++static+string+GetExcelColumnName(long+columnNumber)
++++{
++++++++//max+number+of+column+per+row
++++++++const+long+maxColPerRow+=+702;
++++++++//find+row+number
++++++++long+rowNum+=+(columnNumber+/+maxColPerRow);
++++++++//find+tierable+columns+in+the+row.
++++++++long+dividend+=+columnNumber+-+(maxColPerRow+*+rowNum);

++++++++string+columnName+=+String.Empty;

++++++++long+modulo;

++++++++while+(dividend+>+0)
++++++++{
++++++++++++modulo+=+(dividend+-+1)+%25+26;
++++++++++++columnName+=+Convert.ToChar(65+%2B+modulo).ToString()+%2B+columnName;
++++++++++++dividend+=+(int)((dividend+-+modulo)+/+26);
++++++++}

++++++++return+rowNum%2B1%2B+columnName;
++++}
}|code-block|syntax|javascript|3109208|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is a common question asked in coding test.
it has some constraints:
max columns per row= 702
output should have row number+column name e.g. for 703 answer is 2A.
(note: i have just modified existing code from another answer)
here is the code for the same:

<pre><code> static string GetExcelColumnName(long columnNumber)
 {
 //max number of column per row
 const long maxColPerRow = 702;
 //find row number
 long rowNum = (columnNumber / maxColPerRow);
 //find tierable columns in the row.
 long dividend = columnNumber - (maxColPerRow * rowNum);

 string columnName = String.Empty;

 long modulo;

 while (dividend &gt; 0)
 {
 modulo = (dividend - 1) % 26;
 columnName = Convert.ToChar(65 + modulo).ToString() + columnName;
 dividend = (int)((dividend - modulo) / 26);
 }

 return rowNum+1+ columnName;
 }
}
</code></pre>

blocks|key|807383|text|T-SQL+(SQL+SERVER+18)|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|807384|第一页上的解决方案副本|807385|CREATE+FUNCTION+dbo.getExcelColumnNameByOrdinal(@RowNum+int)++
RETURNS+varchar(5)+++
AS+++
BEGIN++
++++DECLARE+@dividend+int+=+@RowNum;
++++DECLARE+@columnName+varchar(max)+=+'';
++++DECLARE+@modulo+int;

++++WHILE+(@dividend+>+0)
++++BEGIN++
++++++++SELECT+@modulo+=+((@dividend+-+1)+%25+26);
++++++++SELECT+@columnName+=+CHAR((65+%2B+@modulo))+%2B+@columnName;
++++++++SELECT+@dividend+=+CAST(((@dividend+-+@modulo)+/+26)+as+int);
++++END
++++RETURN+
+++++++@columnName;

END;|code-block|syntax|javascript|807386|entityMap^0|0|L|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

T-SQL (SQL SERVER 18)

Copy of the solution on first page

<pre><code>CREATE FUNCTION dbo.getExcelColumnNameByOrdinal(@RowNum int) 
RETURNS varchar(5) 
AS 
BEGIN 
 DECLARE @dividend int = @RowNum;
 DECLARE @columnName varchar(max) = '';
 DECLARE @modulo int;

 WHILE (@dividend &gt; 0)
 BEGIN 
 SELECT @modulo = ((@dividend - 1) % 26);
 SELECT @columnName = CHAR((65 + @modulo)) + @columnName;
 SELECT @dividend = CAST(((@dividend - @modulo) / 26) as int);
 END
 RETURN 
 @columnName;

END;
</code></pre>

blocks|key|3109422|text|下面是基于零的列索引的一个更简单的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3109423|+public+static+string+GetColumnIndexNumberToExcelColumn(int+columnIndex)
++++++++{
++++++++++++int+offset+=+columnIndex+%25+26;
++++++++++++int+multiple+=+columnIndex+/+26;

++++++++++++int+initialSeed+=+65;//Represents+column+"A"
++++++++++++if+(multiple+==+0)
++++++++++++{
++++++++++++++++return+Convert.ToChar(initialSeed+%2B+offset).ToString();
++++++++++++}

++++++++++++return+$"{Convert.ToChar(initialSeed+%2B+multiple+-+1)}{Convert.ToChar(initialSeed+%2B+offset)}";
++++++++}|code-block|syntax|javascript|3109424|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a simpler solution for zero based column Index
<pre><code> public static string GetColumnIndexNumberToExcelColumn(int columnIndex)
 {
 int offset = columnIndex % 26;
 int multiple = columnIndex / 26;

 int initialSeed = 65;//Represents column &quot;A&quot;
 if (multiple == 0)
 {
 return Convert.ToChar(initialSeed + offset).ToString();
 }

 return $&quot;{Convert.ToChar(initialSeed + multiple - 1)}{Convert.ToChar(initialSeed + offset)}&quot;;
 }
</code></pre>

blocks|key|5364005|text|Microsoft+Excel迷你、快速、低俗的配方。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5364006|嗨,|5364007|下面是从数字中获取Excel字符-列-标题的一种方法……|5364008|我为Excel单元格创建了一个公式。|5364009|(也就是说，我采用了不使用VBA编程的方法。)|5364010|该公式查看其中包含数字的单元格，并告诉您该列是什么--用字母表示。|5364011|在附加的图像中：|5364012|我把1,2,3等放在第一行，一直到列ab.，，|5364013|，我把我的公式粘贴到第二行，一直到ab.，|unordered-list-item|5364014|
，|5364015|，我的公式看着第一行，把数字转换成Excel的列标题id。|5364016|
|5364017|我的公式适用于小于702+(Zz)的所有数字。|5364018|5364019|我这样做是为了证明公式有效，这样您就可以查看公式的输出和上面的列标题，并轻松直观地验证公式是否有效。:-)|5364020|5364021|=CONCATENATE(MID("_abcdefghijklmnopqrstuvwxyz"，(IF(MOD(K1,26)>0，INT(K1/26)%2B1，(INT(K1/26)，1)，MID("abcdefghijklmnopqrstuvwxyz"，IF(MOD(K1,26)=0,26，MOD(K1,26)),1))|5364022|下划线在那里是为了调试目的-让你知道有一个实际的空间，它是正确工作的。|5364023|使用上面的公式--无论您放在K1中的是什么--该公式将告诉您列标题将是什么。|5364024|该公式目前的形式仅为两位数(ZZ)，但可以修改为添加第三个字母(ZZZ)。|5364025|​|5364026|📷|atomic|offset|length|5364027|5364028|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/c43695f8cf33f1cbef8ce88e4f40c2fe.png|imageAlt^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|1T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1V|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|1X|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|20|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|21|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|22|8|@]|9|@]|A|$]]|$1|U|3|V|5|R|7|23|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|24|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|R|7|25|8|@]|9|@]|A|$]]|$1|10|3|X|5|6|7|26|8|@]|9|@]|A|$]]|$1|11|3|12|5|R|7|27|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|28|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|29|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|2A|8|@]|9|@]|A|$]]|$1|18|3|19|5|6|7|2B|8|@]|9|@]|A|$]]|$1|1A|3|1B|5|6|7|2C|8|@]|9|@]|A|$]]|$1|1C|3|1D|5|6|7|2D|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|1G|7|2E|8|@]|9|@$1H|2F|1I|2G|1|2H]]|A|$]]|$1|1J|3|1D|5|6|7|2I|8|@]|9|@]|A|$]]|$1|1K|3|-4|5|6|7|2J|8|@]|9|@]|A|$]]]|1L|$1M|$5|1N|1O|1P|A|$1Q|1R|1S|-4]]]]

Microsoft Excel Miniature, Quick-and-Dirty formula.

Hi,

Here's one way to get the Excel character-column-header from a number....

I created a formula for an Excel cell. 

(i.e. I took the approach of not using VBA programming.)

The formula looks at a cell that has a number in it and tells you what the column is -- in letters.

In the attached image:

<ul>
<li>I put 1,2,3 etc in the top row all the way out to column ABS.</li>
<li>I pasted my formula in the second row all the way out to ABS.</li>
<li>My formula looks at row 1 and converts the number to Excel's column header id. </li>
<li>My formula works for all numbers out to 702 (zz).</li>
<li>I did it in this manner to prove that the formula works so you can look at the output from the formula and look at the column header above and easily visually verify that the formula works. :-)

=CONCATENATE(MID("_abcdefghijklmnopqrstuvwxyz",(IF(MOD(K1,26)>0,INT(K1/26)+1,(INT(K1/26)))),1),MID("abcdefghijklmnopqrstuvwxyz",IF(MOD(K1,26)=0,26,MOD(K1,26)),1))</li>
</ul>

The underscore was there for debugging purposes - to let you know there was an actual space and that it was working correctly.

With this formula above -- whatever you put in K1 - the formula will tell you what the column header will be.

The formula, in its current form, only goes out to 2 digits (ZZ) but could be modified to add the 3rd letter (ZZZ).

<a href="https://i.stack.imgur.com/Aqzn4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Aqzn4.png" alt="enter image description here"></a>

blocks|key|3109124|text|下面是我的python解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3109125|import+math

num+=+3500
row_number+=+str(math.ceil(num+/+702))
letters+=+''
num+=+num+-+702+*+math.floor(num+/+702)
while+num:
++++mod+=+(num+-+1)+%25+26
++++letters+%2B=+chr(mod+%2B+65)
++++num+=+(num+-+1)+//+26
result+=+row_number+%2B+("".join(reversed(letters)))
print(result)|code-block|syntax|javascript|3109126|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is my solution in python 

<pre><code>import math

num = 3500
row_number = str(math.ceil(num / 702))
letters = ''
num = num - 702 * math.floor(num / 702)
while num:
 mod = (num - 1) % 26
 letters += chr(mod + 65)
 num = (num - 1) // 26
result = row_number + ("".join(reversed(letters)))
print(result)

</code></pre>

blocks|key|5362928|text|++++public+string+ToBase26(int+number)
++++{
++++++++if+(number+<+0)+return+String.Empty;

++++++++int+remainder+=+number+%25+26;
++++++++int+value+=+number+/+26;

++++++++return+value+==+0+?
++++++++++++String.Format("{0}",+Convert.ToChar(65+%2B+remainder))+:
++++++++++++String.Format("{0}{1}",+ToBase26(value+-+1),+Convert.ToChar(65+%2B+remainder));
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5362929|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> public string ToBase26(int number)
 {
 if (number &lt; 0) return String.Empty;

 int remainder = number % 26;
 int value = number / 26;

 return value == 0 ?
 String.Format("{0}", Convert.ToChar(65 + remainder)) :
 String.Format("{0}{1}", ToBase26(value - 1), Convert.ToChar(65 + remainder));
 }
</code></pre>

How do you convert a numerical number to an Excel column name in C# without using automation getting the value directly from Excel.

Excel 2007 has a possible range of 1 to 16384, which is the number of columns that it supports. The resulting values should be in the form of excel column names, e.g. A, AA, AAA etc.

How to convert a column number (e.g. 127) into an Excel column (e.g. AA)

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

如何在C#中将数值转换为Excel列名，而不使用自动从Excel直接获取值。Excel2007的可能范围是1到16384，这是它支持的列数。结果值应采用excel列名的形式，例如A、AA、AAA等。

问如何将列号(例如127)转换为Excel列(例如AA)
EN

回答 56

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将列号(例如127)转换为Excel列(例如AA)EN

回答 56

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将列号(例如127)转换为Excel列(例如AA)
EN