blocks|key|1855558|text|这基本上就是你需要做的--或者至少，这是最简单的解决方案。你所“浪费”的就是n次方法调用的开销--如果你仔细想想，你实际上不会检查任何情况两次。(IndexOf将在找到匹配项后立即返回，您将继续从它停止的位置继续。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1855559|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|$]]

That's basically what you need to do - or at least, it's the easiest solution. All you'd be "wasting" is the cost of n method invocations - you won't actually be checking any case twice, if you think about it. (IndexOf will return as soon as it finds the match, and you'll keep going from where it left off.)

blocks|key|2069474|text|private+int+IndexOfOccurence(string+s,+string+match,+int+occurence)
{
++++int+i+=+1;
++++int+index+=+0;

++++while+(i+<=+occurence+&&+(index+=+s.IndexOf(match,+index+%2B+1))+!=+-1)
++++{
++++++++if+(i+==+occurence)
++++++++++++return+index;

++++++++i%2B%2B;
++++}

++++return+-1;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2069475|或者在带有扩展方法的C#中|unstyled|2069476|public+static+int+IndexOfOccurence(this+string+s,+string+match,+int+occurence)
{
++++int+i+=+1;
++++int+index+=+0;

++++while+(i+<=+occurence+&&+(index+=+s.IndexOf(match,+index+%2B+1))+!=+-1)
++++{
++++++++if+(i+==+occurence)
++++++++++++return+index;

++++++++i%2B%2B;
++++}

++++return+-1;
}|2069477|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>private int IndexOfOccurence(string s, string match, int occurence)
{
 int i = 1;
 int index = 0;

 while (i &lt;= occurence &amp;&amp; (index = s.IndexOf(match, index + 1)) != -1)
 {
 if (i == occurence)
 return index;

 i++;
 }

 return -1;
}
</code></pre>

or in C# with extension methods

<pre><code>public static int IndexOfOccurence(this string s, string match, int occurence)
{
 int i = 1;
 int index = 0;

 while (i &lt;= occurence &amp;&amp; (index = s.IndexOf(match, index + 1)) != -1)
 {
 if (i == occurence)
 return index;

 i++;
 }

 return -1;
}
</code></pre>

blocks|key|1855634|text|这可能会做到这点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1855635|Console.WriteLine(str.IndexOf((@"\")%2B2)%2B1);|code-block|syntax|javascript|1855636|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 might do it:

<pre><code>Console.WriteLine(str.IndexOf((@"\")+2)+1);
</code></pre>

blocks|key|1855723|text|如果您不需要索引，而是索引处的值，那么使用String.Split()方法并检查请求的匹配项是否在数组中也会更好|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1855724|entityMap^0|L|E|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Maybe it would also be nice to work with the <code>String.Split()</code> Method and check if the requested occurrence is in the array, if you don't need the index, but the value at the index

blocks|key|1855790|text|经过一些基准测试后，这似乎是最简单、最有效的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1855791|public+static+int+IndexOfNthSB(string+input,
+++++++++++++char+value,+int+startIndex,+int+nth)
++++++++{
++++++++++++if+(nth+<+1)
++++++++++++++++throw+new+NotSupportedException("Param+'nth'+must+be+greater+than+0!");
++++++++++++var+nResult+=+0;
++++++++++++for+(int+i+=+startIndex;+i+<+input.Length;+i%2B%2B)
++++++++++++{
++++++++++++++++if+(input[i]+==+value)
++++++++++++++++++++nResult%2B%2B;
++++++++++++++++if+(nResult+==+nth)
++++++++++++++++++++return+i;
++++++++++++}
++++++++++++return+-1;
++++++++}|code-block|syntax|javascript|1855792|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|$]]

After some benchmarking, this seems to be the simplest and most effcient solution

<pre><code>public static int IndexOfNthSB(string input,
 char value, int startIndex, int nth)
 {
 if (nth &lt; 1)
 throw new NotSupportedException("Param 'nth' must be greater than 0!");
 var nResult = 0;
 for (int i = startIndex; i &lt; input.Length; i++)
 {
 if (input[i] == value)
 nResult++;
 if (nResult == nth)
 return i;
 }
 return -1;
 }
</code></pre>

blocks|key|2069774|text|System.ValueTuple+ftw：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2069775|var+index+=+line.Select((x,+i)+=>+(x,+i)).Where(x+=>+x.Item1+==+'"').ElementAt(5).Item2;|offset|length|style|CODE|2069776|从那写一个函数就是家庭作业|2069777|entityMap^0|0|0|2G|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

System.ValueTuple ftw:

<code>var index = line.Select((x, i) =&gt; (x, i)).Where(x =&gt; x.Item1 == '"').ElementAt(5).Item2;</code>

writing a function from that is homework

blocks|key|1855870|text|Tod的答案可以稍微简化一些。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1855871|using+System;

static+class+MainClass+{
++++private+static+int+IndexOfNth(this+string+target,+string+substring,
+++++++++++++++++++++++++++++++++++++++int+seqNr,+int+startIdx+=+0)
++++{
++++++++if+(seqNr+<+1)
++++++++{
++++++++++++throw+new+IndexOutOfRangeException("Parameter+'nth'+must+be+greater+than+0.");
++++++++}

++++++++var+idx+=+target.IndexOf(substring,+startIdx);

++++++++if+(idx+<+0+%7C%7C+seqNr+==+1)+{+return+idx;+}

++++++++return+target.IndexOfNth(substring,+--seqNr,+%2B%2Bidx);+//+skip
++++}

++++static+void+Main+()+{
++++++++Console.WriteLine+("abcbcbcd".IndexOfNth("bc",+1));
++++++++Console.WriteLine+("abcbcbcd".IndexOfNth("bc",+2));
++++++++Console.WriteLine+("abcbcbcd".IndexOfNth("bc",+3));
++++++++Console.WriteLine+("abcbcbcd".IndexOfNth("bc",+4));
++++}
}|code-block|syntax|javascript|1855872|输出|1855873|1
3
5
-1|1855874|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|$]]

Tod's answer can be simplified somewhat.

<pre><code>using System;

static class MainClass {
 private static int IndexOfNth(this string target, string substring,
 int seqNr, int startIdx = 0)
 {
 if (seqNr &lt; 1)
 {
 throw new IndexOutOfRangeException("Parameter 'nth' must be greater than 0.");
 }

 var idx = target.IndexOf(substring, startIdx);

 if (idx &lt; 0 || seqNr == 1) { return idx; }

 return target.IndexOfNth(substring, --seqNr, ++idx); // skip
 }

 static void Main () {
 Console.WriteLine ("abcbcbcd".IndexOfNth("bc", 1));
 Console.WriteLine ("abcbcbcd".IndexOfNth("bc", 2));
 Console.WriteLine ("abcbcbcd".IndexOfNth("bc", 3));
 Console.WriteLine ("abcbcbcd".IndexOfNth("bc", 4));
 }
}
</code></pre>

Output

<pre><code>1
3
5
-1
</code></pre>

blocks|key|2069820|text|或者类似这样的do+while循环|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2069821|+private+static+int+OrdinalIndexOf(string+str,+string+substr,+int+n)
++++{
++++++++int+pos+=+-1;
++++++++do
++++++++{
++++++++++++pos+=+str.IndexOf(substr,+pos+%2B+1);
++++++++}+while+(n--+>+0+&&+pos+!=+-1);
++++++++return+pos;
++++}|code-block|syntax|javascript|2069822|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|$]]

Or something like this with the do while loop

<pre><code> private static int OrdinalIndexOf(string str, string substr, int n)
 {
 int pos = -1;
 do
 {
 pos = str.IndexOf(substr, pos + 1);
 } while (n-- &gt; 0 &amp;&amp; pos != -1);
 return pos;
 }
</code></pre>

Unless I am missing an obvious built-in method, what is the quickest way to get the nth occurrence of a string within a string?

I realize that I could loop the <a href="https://msdn.microsoft.com/en-us/library/k8b1470s(v=vs.110).aspx" rel="noreferrer">IndexOf</a> method by updating its start index on each iteration of the loop. But doing it this way seems wasteful to me.

Get the index of the nth occurrence of a string?

除非我遗漏了一个明显的内置方法，否则获取字符串中字符串的_n_th匹配项的最快方法是什么？我意识到我可以通过在循环的每次迭代中更新方法的开始索引来循环该方法。但这样做对我来说似乎是浪费。

问获取字符串第n次出现的索引？
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