blocks|key|3800087|text|其中39938是自1900年1月1日以来的天数？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3800088|在这种情况下，请使用框架库函数++。|offset|length|style|BOLD|3800089|这个函数encapsulates所有的细节，并做边界检查。|3800090|对于它的历史价值，下面是一个可能的实现：|3800091|(C#)|3800092|public+static+DateTime+FromExcelSerialDate(int+SerialDate)
{
++++if+(SerialDate+>+59)+SerialDate+-=+1;+//Excel/Lotus+2/29/1900+bug+++
++++return+new+DateTime(1899,+12,+31).AddDays(SerialDate);
}|code-block|syntax|javascript|3800093|VB|3800094|Public+Shared+Function+FromExcelSerialDate(ByVal+SerialDate+As+Integer)+As+DateTime
++++If+SerialDate+>+59+Then+SerialDate+-=+1+'+Excel/Lotus+2/29/1900+bug
++++Return+New+DateTime(1899,+12,+31).AddDays(SerialDate)
End+Function|3800095|3800096|更新|3800097|嗯..。一个简单的测试表明它实际上是两天的休息。不知道区别在哪里。|3800098|好的:现在问题解决了。有关详细信息，请参阅评论。|3800099|entityMap|0|LINK|mutability|MUTABLE|url|https://referencesource.microsoft.com/#mscorlib/system/datetime.cs,548^0|0|0|F|H|1|0|4|C|0|0|0|0|0|0|0|0|0|2|0|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1C|8|@$D|1D|E|1E|F|G]|$D|1F|E|1G|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1H|8|@]|9|@$D|1I|E|1J|1|1K]]|A|$]]|$1|J|3|K|5|6|7|1L|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1M|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1N|8|@]|9|@]|A|$Q|R]]|$1|S|3|T|5|6|7|1O|8|@]|9|@]|A|$]]|$1|U|3|V|5|P|7|1P|8|@]|9|@]|A|$Q|R]]|$1|W|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1R|8|@$D|1S|E|1T|F|G]]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1U|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|1V|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]]|14|$15|$5|16|17|18|A|$19|1A]]]]

Where 39938 is the number of days since 1/1/1900?

In that case, use the framework library function <a href="http://msdn.microsoft.com/en-us/library/system.datetime.fromoadate.aspx" rel="noreferrer"><code>DateTime.FromOADate()</code></a> . 
This function <a href="https://referencesource.microsoft.com/#mscorlib/system/datetime.cs,548" rel="noreferrer">encapsulates</a> all the specifics, and does bounds checking.

For its historical value, here is a possible implementation:

(C#)

<pre class="lang-cs prettyprint-override"><code>public static DateTime FromExcelSerialDate(int SerialDate)
{
 if (SerialDate &gt; 59) SerialDate -= 1; //Excel/Lotus 2/29/1900 bug 
 return new DateTime(1899, 12, 31).AddDays(SerialDate);
}
</code></pre>

VB

<pre class="lang-vb prettyprint-override"><code>Public Shared Function FromExcelSerialDate(ByVal SerialDate As Integer) As DateTime
 If SerialDate &gt; 59 Then SerialDate -= 1 ' Excel/Lotus 2/29/1900 bug
 Return New DateTime(1899, 12, 31).AddDays(SerialDate)
End Function
</code></pre>

<hr>

[Update]: 
Hmm... A quick test of that shows it's actually two days off. Not sure where the difference is. 

Okay: problem fixed now. See the comments for details.

blocks|key|1149395|text|我发现使用FromOADate+method更简单，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1149396|DateTime+dt+=+DateTime.FromOADate(39938);|code-block|syntax|javascript|1149397|使用这个代码dt是"05/05/2009“。|1149398|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/system.datetime.fromoadate.aspx^0|5|H|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

I find it simpler using <a href="http://msdn.microsoft.com/en-us/library/system.datetime.fromoadate.aspx" rel="noreferrer">FromOADate method</a>, for example:

<pre><code>DateTime dt = DateTime.FromOADate(39938);
</code></pre>

Using this code dt is "05/05/2009".

blocks|key|1156340|text|对于39938，执行以下操作:+39938+*+864000000000+%2B+599264352000000000|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1156341|864000000000表示一天中的刻度数599264352000000000表示从公元001年到1900年的刻度数|1156342|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

For 39938 do this:
 39938 * 864000000000 + 599264352000000000

864000000000 represents number of ticks in a day
599264352000000000 represents number of ticks from the year 0001 to the year 1900

blocks|key|3800142|text|void+ExcelSerialDateToDMY(int+nSerialDate,+int+&nDay,+
++++++++++++++++++++++++++int+&nMonth,+int+&nYear)
{
++++//+Excel/Lotus+123+have+a+bug+with+29-02-1900.+1900+is+not+a
++++//+leap+year,+but+Excel/Lotus+123+think+it+is...
++++if+(nSerialDate+==+60)
++++{
++++++++nDay++++=+29;
++++++++nMonth++++=+2;
++++++++nYear++++=+1900;

++++++++return;
++++}
++++else+if+(nSerialDate+<+60)
++++{
++++++++//+Because+of+the+29-02-1900+bug,+any+serial+date+
++++++++//+under+60+is+one+off...+Compensate.
++++++++nSerialDate%2B%2B;
++++}

++++//+Modified+Julian+to+DMY+calculation+with+an+addition+of+2415019
++++int+l+=+nSerialDate+%2B+68569+%2B+2415019;
++++int+n+=+int((+4+*+l+)+/+146097);
++++++++++++l+=+l+-+int((+146097+*+n+%2B+3+)+/+4);
++++int+i+=+int((+4000+*+(+l+%2B+1+)+)+/+1461001);
++++++++l+=+l+-+int((+1461+*+i+)+/+4)+%2B+31;
++++int+j+=+int((+80+*+l+)+/+2447);
+++++nDay+=+l+-+int((+2447+*+j+)+/+80);
++++++++l+=+int(j+/+11);
++++++++nMonth+=+j+%2B+2+-+(+12+*+l+);
++++nYear+=+100+*+(+n+-+49+)+%2B+i+%2B+l;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3800143|剪切和粘贴某人的才华。|unstyled|3800144|Ian+Brown|offset|length|3800145|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ianbrownphotography.co.uk/^0|0|0|0|9|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>void ExcelSerialDateToDMY(int nSerialDate, int &amp;nDay, 
 int &amp;nMonth, int &amp;nYear)
{
 // Excel/Lotus 123 have a bug with 29-02-1900. 1900 is not a
 // leap year, but Excel/Lotus 123 think it is...
 if (nSerialDate == 60)
 {
 nDay = 29;
 nMonth = 2;
 nYear = 1900;

 return;
 }
 else if (nSerialDate &lt; 60)
 {
 // Because of the 29-02-1900 bug, any serial date 
 // under 60 is one off... Compensate.
 nSerialDate++;
 }

 // Modified Julian to DMY calculation with an addition of 2415019
 int l = nSerialDate + 68569 + 2415019;
 int n = int(( 4 * l ) / 146097);
 l = l - int(( 146097 * n + 3 ) / 4);
 int i = int(( 4000 * ( l + 1 ) ) / 1461001);
 l = l - int(( 1461 * i ) / 4) + 31;
 int j = int(( 80 * l ) / 2447);
 nDay = l - int(( 2447 * j ) / 80);
 l = int(j / 11);
 nMonth = j + 2 - ( 12 * l );
 nYear = 100 * ( n - 49 ) + i + l;
}
</code></pre>

Cut and Paste of someone elses talents...

<a href="http://www.ianbrownphotography.co.uk" rel="nofollow noreferrer">Ian Brown</a>

How do I convert from excel serial date to a .NET date time?

For example 39938 is 05/05/2009.

How do I convert an Excel serial date number to a .NET DateTime?

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。

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如何将excel序列日期转换为.NET日期时间？例如39938是05/05/2009.

问如何将Excel序列号转换为.NET DateTime？
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问如何将Excel序列号转换为.NET DateTime？
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