blocks|key|3229799|text|出现两个简单的选项：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3229800|3229801|Use+charAt()+repeatedly|unordered-list-item|offset|length|style|CODE|3229802|Use+indexOf()+repeatedly|3229803|3229804|例如：|3229805|public+static+int+nthIndexOf(String+text,+char+needle,+int+n)
{
++++for+(int+i+=+0;+i+<+text.length();+i%2B%2B)
++++{
++++++++if+(text.charAt(i)+==+needle)
++++++++{
++++++++++++n--;
++++++++++++if+(n+==+0)
++++++++++++{
++++++++++++++++return+i;
++++++++++++}
++++++++}
++++}
++++return+-1;
}|code-block|syntax|javascript|3229806|这可能不如反复使用indexOf那么好，但可能更简单。|3229807|entityMap^0|0|0|4|8|0|4|9|0|0|0|0|9|7|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|Z|8|@$F|10|G|11|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|12|8|@$F|13|G|14|H|I]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|17|8|@]|9|@]|A|$R|S]]|$1|T|3|U|5|6|7|18|8|@$F|19|G|1A|H|I]]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|W|$]]

Two simple options occur:

<ul>
<li>Use <code>charAt()</code> repeatedly</li>
<li>Use <code>indexOf()</code> repeatedly</li>
</ul>

For example:

<pre><code>public static int nthIndexOf(String text, char needle, int n)
{
 for (int i = 0; i &lt; text.length(); i++)
 {
 if (text.charAt(i) == needle)
 {
 n--;
 if (n == 0)
 {
 return i;
 }
 }
 }
 return -1;
}
</code></pre>

That may well not perform as well as using <code>indexOf</code> repeatedly, but it's possibly simpler to get right.

blocks|key|3229811|text|+([.%5E/]*/){2}[%5E/]*(/)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3229812|匹配任何后跟/的内容两次，然后再次匹配。第三个是你想要的那个|unstyled|3229813|Matcher状态可以用来告诉我们最后一个/在哪里|offset|length|3229814|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/1.4.2/docs/api/java/util/regex/Matcher.html^0|0|0|0|7|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> ([.^/]*/){2}[^/]*(/)
</code></pre>

Match anything followed by / two times, then again. The third one is the one you want

The <a href="http://download.oracle.com/javase/1.4.2/docs/api/java/util/regex/Matcher.html" rel="noreferrer">Matcher</a> state can be used to tell where the last / is

blocks|key|296249|text|如果你的项目已经依赖于Apache+Commons，你可以使用StringUtils.ordinalIndexOf，否则，这里有一个实现：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|296250|public+static+int+ordinalIndexOf(String+str,+String+substr,+int+n)+{
++++int+pos+=+str.indexOf(substr);
++++while+(--n+>+0+&&+pos+!=+-1)
++++++++pos+=+str.indexOf(substr,+pos+%2B+1);
++++return+pos;
}|code-block|syntax|javascript|296251|296252|这篇文章已经被重写为一篇文章here。|296253|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/StringUtils.html#ordinalIndexOf%2528java.lang.CharSequence,%2520java.lang.CharSequence,%2520int%2529|1|http://programming.guide/java/nth-occurrence-in-string.html^0|V|Q|V|Q|0|0|0|0|E|4|1|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@$9|10|A|11|1|12]]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|14|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|N|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]|V|$5|Q|R|S|E|$T|W]]]]

If your project already depends on Apache Commons you can use <a href="https://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/StringUtils.html#ordinalIndexOf%28java.lang.CharSequence,%20java.lang.CharSequence,%20int%29" rel="noreferrer"><code>StringUtils.ordinalIndexOf</code></a>, otherwise, here's an implementation:

<pre><code>public static int ordinalIndexOf(String str, String substr, int n) {
 int pos = str.indexOf(substr);
 while (--n &gt; 0 &amp;&amp; pos != -1)
 pos = str.indexOf(substr, pos + 1);
 return pos;
}
</code></pre>

<hr>

This post has been rewritten as an article <a href="http://programming.guide/java/nth-occurrence-in-string.html" rel="noreferrer">here</a>.

blocks|key|3229900|text|public+static+int+nth(String+source,+String+pattern,+int+n)+{

+++int+i+=+0,+pos+=+0,+tpos+=+0;

+++while+(i+<+n)+{

++++++pos+=+source.indexOf(pattern);
++++++if+(pos+>+-1)+{
+++++++++source+=+source.substring(pos%2B1);
+++++++++tpos+%2B=+pos%2B1;
+++++++++i%2B%2B;
++++++}+else+{
+++++++++return+-1;
++++++}
+++}

+++return+tpos+-+1;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3229901|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 int nth(String source, String pattern, int n) {

 int i = 0, pos = 0, tpos = 0;

 while (i &lt; n) {

 pos = source.indexOf(pattern);
 if (pos &gt; -1) {
 source = source.substring(pos+1);
 tpos += pos+1;
 i++;
 } else {
 return -1;
 }
 }

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

blocks|key|3229994|text|您可以尝试如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3229995|import+java.util.regex.Matcher;
import+java.util.regex.Pattern;

public+class+Main+{
++++public+static+void+main(String[]+args)+{
++++++System.out.println(from3rd("/folder1/folder2/folder3/"));
++++}

++++private+static+Pattern+p+=+Pattern.compile("(/[%5E/]*){2}/([%5E/]*)");

++++public+static+String+from3rd(String+in)+{
++++++++Matcher+m+=+p.matcher(in);

++++++++if+(m.matches())
++++++++++++return+m.group(2);
++++++++else
++++++++++++return+null;
++++}
}|code-block|syntax|javascript|3229996|请注意，我在regex中做了一些假设：|3229997|3229998|输入路径是绝对路径(即以"/“开头)；|unordered-list-item|3229999|结果中不需要第三个”/“。|3230000|3230001|3230002|按照评论中的要求，我将尝试解释正则表达式：(/[%5E/]*){2}/([%5E/]*)|offset|length|style|CODE|3230003|​|3230004|📷|atomic|3230005|3230006|3230007|/[%5E/]*是一个后跟|3230008|(/[%5E/]*)的/+(不是[%5E/]*的任意数量的字符)，|3230009|(/[%5E/]*)将前一个表达式分组到单个实体中。这是表达式的第st组，|3230010|(/[%5E/]*){2}表示该组必须精确匹配{2}时间，|3230011|([%5E/]*)也是任意数量的不是1的字符，{2}将前一个表达式分组在单个实体中。这是表达式的2nd组。|3230012|3230013|这样，您只需获取与第二组匹配的子字符串：return+m.group(2);|3230014|图片由Debuggex提供|3230015|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://www.debuggex.com/i/2sAKhlPpasMkBR31.png|imageAlt|1|LINK|MUTABLE|url|https://www.debuggex.com/r/2sAKhlPpasMkBR31^0|0|0|0|0|0|0|0|0|L|J|0|0|0|1|0|0|0|0|0|6|0|0|8|9|1|E|5|0|0|8|0|0|B|L|3|0|0|7|G|1|L|3|1A|1|0|0|K|I|0|3|8|1|0^^$0|@$1|2|3|4|5|6|7|1W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|20|8|@]|9|@]|A|$]]|$1|M|3|N|5|L|7|21|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|22|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|23|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|24|8|@$S|25|T|26|U|V]]|9|@]|A|$]]|$1|W|3|X|5|6|7|27|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|10|7|28|8|@]|9|@$S|29|T|2A|1|2B]]|A|$]]|$1|11|3|X|5|6|7|2C|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|2D|8|@]|9|@]|A|$]]|$1|13|3|14|5|L|7|2E|8|@$S|2F|T|2G|U|V]]|9|@]|A|$]]|$1|15|3|16|5|L|7|2H|8|@$S|2I|T|2J|U|V]|$S|2K|T|2L|U|V]|$S|2M|T|2N|U|V]]|9|@]|A|$]]|$1|17|3|18|5|L|7|2O|8|@$S|2P|T|2Q|U|V]]|9|@]|A|$]]|$1|19|3|1A|5|L|7|2R|8|@$S|2S|T|2T|U|V]|$S|2U|T|2V|U|V]]|9|@]|A|$]]|$1|1B|3|1C|5|L|7|2W|8|@$S|2X|T|2Y|U|V]|$S|2Z|T|30|U|V]|$S|31|T|32|U|V]|$S|33|T|34|U|V]]|9|@]|A|$]]|$1|1D|3|-4|5|6|7|35|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|6|7|36|8|@$S|37|T|38|U|V]]|9|@]|A|$]]|$1|1G|3|1H|5|6|7|39|8|@]|9|@$S|3A|T|3B|1|3C]]|A|$]]|$1|1I|3|-4|5|6|7|3D|8|@]|9|@]|A|$]]]|1J|$1K|$5|1L|1M|1N|A|$1O|1P|1Q|-4]]|1R|$5|1S|1M|1T|A|$1U|1V]]]]

You can try something like this:

<pre><code>import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {
 public static void main(String[] args) {
 System.out.println(from3rd("/folder1/folder2/folder3/"));
 }

 private static Pattern p = Pattern.compile("(/[^/]*){2}/([^/]*)");

 public static String from3rd(String in) {
 Matcher m = p.matcher(in);

 if (m.matches())
 return m.group(2);
 else
 return null;
 }
}
</code></pre>

Note that I did some assumptions in the regex:

<ul>
<li>the input path is absolute (i.e. starts with "/");</li>
<li>you do not need the 3rd "/" in the result.</li>
</ul>

<hr>

As requested in a comment, I'll try to explain the regex: <code>(/[^/]*){2}/([^/]*)</code>

<img src="https://www.debuggex.com/i/2sAKhlPpasMkBR31.png" alt="Regular expression visualization">

<ul>
<li><code>/[^/]*</code> is a <code>/</code> followed by <code>[^/]*</code> (any number of characters that are not a <code>/</code>),</li>
<li><code>(/[^/]*)</code> groups the previous expression in a single entity. This is the <code>1</code>st group of the expression,</li>
<li><code>(/[^/]*){2}</code> means that the group must match extactly <code>{2}</code> times,</li>
<li><code>[^/]*</code> is again any number of characters that are not a <code>/</code>,</li>
<li><code>([^/]*)</code> groups the previos expression in a single entity. This is the <code>2</code>nd group of the expression.</li>
</ul>

This way you have only to get the substring that matches the 2nd group: <code>return m.group(2);</code>

Image courtesy by <a href="https://www.debuggex.com/r/2sAKhlPpasMkBR31" rel="noreferrer">Debuggex</a>

blocks|key|293610|text|另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|293611|public+static+void+main(String[]+args)+{
++++String+str+=+"/folder1/folder2/folder3/";+
++++int+index+=+nthOccurrence(str,+'/',+3);
++++System.out.println(index);
}

public+static+int+nthOccurrence(String+s,+char+c,+int+occurrence)+{
++++return+nthOccurrence(s,+0,+c,+0,+occurrence);
}

public+static+int+nthOccurrence(String+s,+int+from,+char+c,+int+curr,+int+expected)+{
++++final+int+index+=+s.indexOf(c,+from);
++++if(index+==+-1)+return+-1;
++++return+(curr+%2B+1+==+expected)+?+index+:+
++++++++nthOccurrence(s,+index+%2B+1,+c,+curr+%2B+1,+expected);
}|code-block|syntax|javascript|293612|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 approach:

<pre><code>public static void main(String[] args) {
 String str = "/folder1/folder2/folder3/"; 
 int index = nthOccurrence(str, '/', 3);
 System.out.println(index);
}

public static int nthOccurrence(String s, char c, int occurrence) {
 return nthOccurrence(s, 0, c, 0, occurrence);
}

public static int nthOccurrence(String s, int from, char c, int curr, int expected) {
 final int index = s.indexOf(c, from);
 if(index == -1) return -1;
 return (curr + 1 == expected) ? index : 
 nthOccurrence(s, index + 1, c, curr + 1, expected);
}
</code></pre>

blocks|key|3230035|text|我认为查找字符串第N次出现的最简单的解决方案是使用Apache+Commons中的StringUtils.ordinalIndexOf()。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3230036|示例：|3230037|StringUtils.ordinalIndexOf("aabaabaa",+"b",+2)++==+5|code-block|syntax|javascript|3230038|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#ordinalIndexOf%2528java.lang.String,%2520java.lang.String,%2520int%2529^0|15|S|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

I believe the easiest solution for finding the Nth occurrence of a String is to use <a href="http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#ordinalIndexOf%28java.lang.String,%20java.lang.String,%20int%29">StringUtils.ordinalIndexOf()</a> from Apache Commons.

Example:

<pre><code>StringUtils.ordinalIndexOf("aabaabaa", "b", 2) == 5
</code></pre>

blocks|key|294404|text|我对aioobe的答案做了一些修改，得到了第n个lastIndexOf版本，并修复了一些NPE问题。请看下面的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|294405|public+int+nthLastIndexOf(String+str,+char+c,+int+n)+{
++++++++if+(str+==+null+%7C%7C+n+<+1)
++++++++++++return+-1;
++++++++int+pos+=+str.length();
++++++++while+(n--+>+0+&&+pos+!=+-1)
++++++++++++pos+=+str.lastIndexOf(c,+pos+-+1);
++++++++return+pos;
}|code-block|syntax|javascript|294406|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 made a few changes to aioobe's answer and got a nth lastIndexOf version, and fix some NPE problems. See code below:

<pre><code>public int nthLastIndexOf(String str, char c, int n) {
 if (str == null || n &lt; 1)
 return -1;
 int pos = str.length();
 while (n-- &gt; 0 &amp;&amp; pos != -1)
 pos = str.lastIndexOf(c, pos - 1);
 return pos;
}
</code></pre>

blocks|key|291348|text|/*+program+to+find+nth+occurence+of+a+character+*/

import+java.util.Scanner;

public+class+CharOccur1
{

++++public+static+void+main(String+arg[])
++++{
++++++++Scanner+scr=new+Scanner(System.in);
++++++++int+position=-1,count=0;
++++++++System.out.println("enter+the+string");
++++++++String+str=scr.nextLine();
++++++++System.out.println("enter+the+nth+occurence+of+the+character");
++++++++int+n=Integer.parseInt(scr.next());
++++++++int+leng=str.length();
++++++++char+c[]=new+char[leng];
++++++++System.out.println("Enter+the+character+to+find");
++++++++char+key=scr.next().charAt(0);
++++++++c=str.toCharArray();
++++++++for(int+i=0;i<c.length;i%2B%2B)
++++++++{
++++++++++++if(c[i]==key)
++++++++++++{
++++++++++++++++count%2B%2B;
++++++++++++++++position=i;
++++++++++++++++if(count==n)
++++++++++++++++{
++++++++++++++++++++System.out.println("Character+found");
++++++++++++++++++++System.out.println("the+position+at+which+the+"+%2B+count+%2B+"+ocurrence+occurs+is+"+%2B+position);
++++++++++++++++++++return;
++++++++++++++++}
++++++++++++}
++++++++}
++++++++if(n>count)
++++++++{+
++++++++++++System.out.println("Character+occurs++"%2B+count+%2B+"+times");
++++++++++++return;
++++++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|291349|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>/* program to find nth occurence of a character */

import java.util.Scanner;

public class CharOccur1
{

 public static void main(String arg[])
 {
 Scanner scr=new Scanner(System.in);
 int position=-1,count=0;
 System.out.println("enter the string");
 String str=scr.nextLine();
 System.out.println("enter the nth occurence of the character");
 int n=Integer.parseInt(scr.next());
 int leng=str.length();
 char c[]=new char[leng];
 System.out.println("Enter the character to find");
 char key=scr.next().charAt(0);
 c=str.toCharArray();
 for(int i=0;i&lt;c.length;i++)
 {
 if(c[i]==key)
 {
 count++;
 position=i;
 if(count==n)
 {
 System.out.println("Character found");
 System.out.println("the position at which the " + count + " ocurrence occurs is " + position);
 return;
 }
 }
 }
 if(n&gt;count)
 { 
 System.out.println("Character occurs "+ count + " times");
 return;
 }
 }
}
</code></pre>

blocks|key|294416|text|现在有了Apache+Commons+Lang的StringUtils支持，|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|294417|这是原语：|style|BOLD|294418|int+org.apache.commons.lang.StringUtils.ordinalIndexOf(CharSequence+str,+CharSequence+searchStr,+int+ordinal)|code-block|syntax|javascript|294419|对于您的问题，您可以编写以下代码：StringUtils.ordinalIndexOf(uri,+"/",+3)|CODE|294420|还可以使用lastOrdinalIndexOf方法查找字符串中字符的最后第n个匹配项。|294421|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#ordinalIndexOf%2528java.lang.String,%2520java.lang.String,%2520int%2529|1|http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#lastOrdinalIndexOf%2528java.lang.String,%2520java.lang.String,%2520int%2529^0|O|B|0|0|0|5|0|0|H|13|0|5|I|1|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|D|3|E|5|6|7|15|8|@$A|16|B|17|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|18|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|19|8|@$A|1A|B|1B|F|O]]|9|@]|C|$]]|$1|P|3|Q|5|6|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|R|3|-4|5|6|7|1G|8|@]|9|@]|C|$]]]|S|$T|$5|U|V|W|C|$X|Y]]|Z|$5|U|V|W|C|$X|10]]]]

Nowadays there IS support of Apache Commons Lang's <a href="http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#ordinalIndexOf%28java.lang.String,%20java.lang.String,%20int%29" rel="nofollow">StringUtils</a>,

This is the primitive:

<pre><code>int org.apache.commons.lang.StringUtils.ordinalIndexOf(CharSequence str, CharSequence searchStr, int ordinal)
</code></pre>

for your problem you can code the following: <code>StringUtils.ordinalIndexOf(uri, "/", 3)</code>

You can also find the last nth occurrence of a character in a string with the <a href="http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#lastOrdinalIndexOf%28java.lang.String,%20java.lang.String,%20int%29" rel="nofollow">lastOrdinalIndexOf</a> method.

blocks|key|294423|text|public+class+Sam_Stringnth+{

++++public+static+void+main(String[]+args)+{
++++++++String+str="abcabcabc";
++++++++int+n+=+nthsearch(str,+'c',+3);
++++++++if(n<=0)
++++++++++++System.out.println("Character+not+found");
++++++++else
++++++++++++System.out.println("Position+is:"%2Bn);
++++}
++++public+static+int+nthsearch(String+str,+char+ch,+int+n){
++++++++int+pos=0;
++++++++if(n!=0){
++++++++++++for(int+i=1;+i<=n;i%2B%2B){
++++++++++++++++pos+=+str.indexOf(ch,+pos)%2B1;
++++++++++++}
++++++++++++return+pos;
++++++++}
++++++++else{
++++++++++++return+0;
++++++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|294424|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 class Sam_Stringnth {

 public static void main(String[] args) {
 String str="abcabcabc";
 int n = nthsearch(str, 'c', 3);
 if(n&lt;=0)
 System.out.println("Character not found");
 else
 System.out.println("Position is:"+n);
 }
 public static int nthsearch(String str, char ch, int n){
 int pos=0;
 if(n!=0){
 for(int i=1; i&lt;=n;i++){
 pos = str.indexOf(ch, pos)+1;
 }
 return pos;
 }
 else{
 return 0;
 }
 }
}
</code></pre>

blocks|key|3230198|text|这个答案改进了@aioobe的答案，修复了答案中的两个bug。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3230199|3230200|n=0应返回-1。|ordered-list-item|3230201|第n次出现返回-1，但它在第n-1次出现时有效。|3230202|3230203|试试这个！|3230204|++++public+int+nthOccurrence(String+str,+char+c,+int+n)+{
++++if(n+<=+0){
++++++++return+-1;
++++}
++++int+pos+=+str.indexOf(c,+0);
++++while+(n--+>+1+&&+pos+!=+-1)
++++++++pos+=+str.indexOf(c,+pos%2B1);
++++return+pos;
}|code-block|syntax|javascript|3230205|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|S|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|T|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|U|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|W|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|X|8|@]|9|@]|A|$N|O]]|$1|P|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|Q|$]]

This answer improves on @aioobe 's answer. Two bugs in that answer were fixed. 
1. n=0 should return -1. 
2. nth occurence returned -1, but it worked on n-1th occurences.

Try this !

<pre><code> public int nthOccurrence(String str, char c, int n) {
 if(n &lt;= 0){
 return -1;
 }
 int pos = str.indexOf(c, 0);
 while (n-- &gt; 1 &amp;&amp; pos != -1)
 pos = str.indexOf(c, pos+1);
 return pos;
}
</code></pre>

blocks|key|291372|text|我的解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|291373|/**
+*+Like+String.indexOf,+but+find+the+n:th+occurance+of+c
+*+@param+s+string+to+search
+*+@param+c+character+to+search+for
+*+@param+n+n:th+character+to+seach+for,+starting+with+1
+*+@return+the+position+(0-based)+of+the+found+char,+or+-1+if+failed
+*/

public+static+int+nthIndexOf(String+s,+char+c,+int+n)+{
++++int+i+=+-1;
++++while+(n--+>+0)+{
++++++++i+=+s.indexOf(c,+i+%2B+1);
++++++++if+(i+==+-1)
++++++++++++break;
++++}
++++return+i;
}|code-block|syntax|javascript|291374|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|$]]

My solution:

<pre><code>/**
 * Like String.indexOf, but find the n:th occurance of c
 * @param s string to search
 * @param c character to search for
 * @param n n:th character to seach for, starting with 1
 * @return the position (0-based) of the found char, or -1 if failed
 */

public static int nthIndexOf(String s, char c, int n) {
 int i = -1;
 while (n-- &gt; 0) {
 i = s.indexOf(c, i + 1);
 if (i == -1)
 break;
 }
 return i;
}
</code></pre>

blocks|key|3230230|text|该代码返回第n个出现位置的子字符串，即字段宽度。举例说明。如果字符串“Stack+overflow+in+low+melow"是搜索令牌"low”的第二次和第二次出现的字符串，你会同意我的说法，它第二次出现的位置是"18++21"。indexOfOccurance("Stack+overflow+in+low+melow"，low，2)返回字符串中的18和21。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3230231|class+Example{
++++public+Example(){
++++}
++++++++++++public+String+indexOfOccurance(String+string,+String+token,+int+nthOccurance)+{
++++++++++++++++++++int+lengthOfToken+=+token.length();
++++++++++++++++++++int+nthCount+=+0;
++++++++++++++++++++for+(int+shift+=+0,count+=+0;+count+<+string.length()+-+token.length()+%2B+2;+count%2B%2B,+shift%2B%2B,+lengthOfToken%2B%2B)
++++++++++++++++++++++++if+(string.substring(shift,+lengthOfToken).equalsIgnoreCase(token))+{+
++++++++++++++++++++//+keeps+count+of+nthOccurance
++++++++++++++++++++++++++++nthCount%2B%2B;+
++++++++++++++++++++++++if+(nthCount+==+nthOccurance){
++++++++++++++++++++//checks+if+nthCount++==+nthOccurance.+If+true,+then+breaks+
+++++++++++++++++++++++++++++return+String.valueOf(shift)%2B+"+"+%2BString.valueOf(lengthOfToken);+++
++++++++++++++++++++++++}++
++++++++++++++++++++}
++++++++++++++++++++return+"-1";
++++++++++++++++}
++++public+static+void+main(String+args[]){
++++Example+example+=+new+Example();
++++String+string+=+"the+man,+the+woman+and+the+child";
++++int+nthPositionOfThe+=+3;
+++System.out.println("3rd+Occurance+of+the+is+at+"+%2B+example.indexOfOccurance(string,+"the",+nthPositionOfThe));
++++}
++++}|code-block|syntax|javascript|3230232|entityMap^0|2Z|23|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|$]]

The code returns the nth occurrence positions substring aka field width. Example. if string "Stack overflow in low melow" is the string to search 2nd occurance of token "low", you will agree with me that it 2nd occurance is at subtring "18 and 21". indexOfOccurance("Stack overflow in low melow", low, 2) returns 18 and 21 in a string.

<pre><code>class Example{
 public Example(){
 }
 public String indexOfOccurance(String string, String token, int nthOccurance) {
 int lengthOfToken = token.length();
 int nthCount = 0;
 for (int shift = 0,count = 0; count &lt; string.length() - token.length() + 2; count++, shift++, lengthOfToken++)
 if (string.substring(shift, lengthOfToken).equalsIgnoreCase(token)) { 
 // keeps count of nthOccurance
 nthCount++; 
 if (nthCount == nthOccurance){
 //checks if nthCount == nthOccurance. If true, then breaks 
 return String.valueOf(shift)+ " " +String.valueOf(lengthOfToken); 
 } 
 }
 return "-1";
 }
 public static void main(String args[]){
 Example example = new Example();
 String string = "the man, the woman and the child";
 int nthPositionOfThe = 3;
 System.out.println("3rd Occurance of the is at " + example.indexOfOccurance(string, "the", nthPositionOfThe));
 }
 }
</code></pre>

blocks|key|3230238|text|也许您可以通过String.split(..)实现这一点。方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3230239|String+str+=+"";
String[]+tokens+=+str.split("/")
return+tokens[nthIndex]+==+null+|code-block|syntax|javascript|3230240|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|$]]

May be you could achieve this through String.split(..) method also.

<pre><code>String str = "";
String[] tokens = str.split("/")
return tokens[nthIndex] == null 
</code></pre>

blocks|key|3015991|text|public+static+int+findNthOccurrence(String+phrase,+String+str,+int+n)
{
++++int+val+=+0,+loc+=+-1;
++++for(int+i+=+0;+i+<=+phrase.length()-str.length()+&&+val+<+n;+i%2B%2B)
++++{
++++++++if(str.equals(phrase.substring(i,i%2Bstr.length())))
++++++++{
++++++++++++val%2B%2B;
++++++++++++loc+=+i;
++++++++}
++++}

++++if(val+==+n)
++++++++return+loc;
++++else
++++++++return+-1;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3015992|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 int findNthOccurrence(String phrase, String str, int n)
{
 int val = 0, loc = -1;
 for(int i = 0; i &lt;= phrase.length()-str.length() &amp;&amp; val &lt; n; i++)
 {
 if(str.equals(phrase.substring(i,i+str.length())))
 {
 val++;
 loc = i;
 }
 }

 if(val == n)
 return loc;
 else
 return -1;
}
</code></pre>

blocks|key|3016009|text|//scala|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3016010|//如果该值在第n次不存在，则抛出-1，即使它一直存在到第n-1次。//如果值出现第n次，则抛出的索引|3016011|def+indexOfWithNumber(tempString:String,valueString:String,numberOfOccurance:Int):Int={
var+stabilizeIndex=0+
var+tempSubString=tempString+
var+tempIndex=tempString.indexOf(valueString)+
breakable
{
for+(+i+<-+1+to+numberOfOccurance)
if+((tempSubString.indexOf(valueString)+!=+-1)+&&+(tempIndex+!=+-1))
{
tempIndex=tempSubString.indexOf(valueString)
tempSubString=tempSubString.substring(tempIndex%2B1,tempSubString.size)+//+ADJUSTING+FOR+0
stabilizeIndex=stabilizeIndex%2BtempIndex%2B1+//+ADJUSTING+FOR+0
}
else
{+
stabilizeIndex=+-1
tempIndex=+0
break
}
}
stabilizeIndex+match+{+case+value+if+value+<=+-1+=>+-1+case+_+=>+stabilizeIndex-1+}+//+reverting+for+adjusting+0+previously
}


indexOfWithNumber("bbcfgtbgft","b",3)+//+6
indexOfWithNumber("bbcfgtbgft","b",2)+//1
indexOfWithNumber("bbcfgtbgft","b",4)+//-1

indexOfWithNumber("bbcfgtbcgft","bc",1)++//1
indexOfWithNumber("bbcfgtbcgft","bc",4)+//-1
indexOfWithNumber("bbcfgtbcgft","bc",2)+//6|code-block|syntax|javascript|3016012|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|$]]

//scala
// throw's -1 if the value isn't present for nth time, even if it is present till n-1 th time.
// throw's index if the value is present for nth time
<pre><code>def indexOfWithNumber(tempString:String,valueString:String,numberOfOccurance:Int):Int={
var stabilizeIndex=0 
var tempSubString=tempString 
var tempIndex=tempString.indexOf(valueString) 
breakable
{
for ( i &lt;- 1 to numberOfOccurance)
if ((tempSubString.indexOf(valueString) != -1) &amp;&amp; (tempIndex != -1))
{
tempIndex=tempSubString.indexOf(valueString)
tempSubString=tempSubString.substring(tempIndex+1,tempSubString.size) // ADJUSTING FOR 0
stabilizeIndex=stabilizeIndex+tempIndex+1 // ADJUSTING FOR 0
}
else
{ 
stabilizeIndex= -1
tempIndex= 0
break
}
}
stabilizeIndex match { case value if value &lt;= -1 =&gt; -1 case _ =&gt; stabilizeIndex-1 } // reverting for adjusting 0 previously
}


indexOfWithNumber(&quot;bbcfgtbgft&quot;,&quot;b&quot;,3) // 6
indexOfWithNumber(&quot;bbcfgtbgft&quot;,&quot;b&quot;,2) //1
indexOfWithNumber(&quot;bbcfgtbgft&quot;,&quot;b&quot;,4) //-1

indexOfWithNumber(&quot;bbcfgtbcgft&quot;,&quot;bc&quot;,1) //1
indexOfWithNumber(&quot;bbcfgtbcgft&quot;,&quot;bc&quot;,4) //-1
indexOfWithNumber(&quot;bbcfgtbcgft&quot;,&quot;bc&quot;,2) //6
</code></pre>

blocks|key|293694|text|APACHE实现(复制-粘贴:不要为一个函数导入整个库！)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|293695|下面是与其StringUtils库解耦的Apache+Commons实现(这样您只需复制粘贴此，而不必为一个函数添加库的依赖项)：|offset|length|style|BOLD|293696|/**
+*+Finds+the+n-th+index+within+a+String,+handling+{@code+null}.
+*+This+method+uses+{@link+String#indexOf(String)}+if+possible.
+*+Note+that+matches+may+overlap
+*
+*+A+{@code+null}+CharSequence+will+return+{@code+-1}.
+*
+*+@param+str++the+CharSequence+to+check,+may+be+null
+*+@param+searchStr++the+CharSequence+to+find,+may+be+null
+*+@param+ordinal++the+n-th+{@code+searchStr}+to+find,+overlapping+matches+are+allowed.
+*+@param+lastIndex+true+if+lastOrdinalIndexOf()+otherwise+false+if+ordinalIndexOf()
+*+@return+the+n-th+index+of+the+search+CharSequence,
+*++{@code+-1}+if+no+match+or+{@code+null}+string+input
+*/
private+static+int+ordinalIndexOf(final+String+str,+final+String+searchStr,+final+int+ordinal,+final+boolean+lastIndex)+{
++++if+(str+==+null+%7C%7C+searchStr+==+null+%7C%7C+ordinal+<=+0)+{
++++++++return+-1;
++++}
++++if+(searchStr.length()+==+0)+{
++++++++return+lastIndex+?+str.length()+:+0;
++++}
++++int+found+=+0;
++++//+set+the+initial+index+beyond+the+end+of+the+string
++++//+this+is+to+allow+for+the+initial+index+decrement/increment
++++int+index+=+lastIndex+?+str.length()+:+-1;
++++do+{
++++++++if+(lastIndex)+{
++++++++++++index+=+str.lastIndexOf(searchStr,+index+-+1);+//+step+backwards+thru+string
++++++++}+else+{
++++++++++++index+=+str.indexOf(searchStr,+index+%2B+1);+//+step+forwards+through+string
++++++++}
++++++++if+(index+<+0)+{
++++++++++++return+index;
++++++++}
++++++++found%2B%2B;
++++}+while+(found+<+ordinal);
++++return+index;
}|code-block|syntax|javascript|293697|entityMap^0|0|1B|9|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

<h2>APACHE IMPLEMENTATION (copy-paste: don't import a whole library for one function!)</h2>
Here is the exact Apache Commons implementation decoupled from their StringUtils library (so that you can just copy paste this and don't have to add a dependency for the library for just one function):
<pre><code>/**
 * &lt;p&gt;Finds the n-th index within a String, handling {@code null}.
 * This method uses {@link String#indexOf(String)} if possible.&lt;/p&gt;
 * &lt;p&gt;Note that matches may overlap&lt;p&gt;
 *
 * &lt;p&gt;A {@code null} CharSequence will return {@code -1}.&lt;/p&gt;
 *
 * @param str the CharSequence to check, may be null
 * @param searchStr the CharSequence to find, may be null
 * @param ordinal the n-th {@code searchStr} to find, overlapping matches are allowed.
 * @param lastIndex true if lastOrdinalIndexOf() otherwise false if ordinalIndexOf()
 * @return the n-th index of the search CharSequence,
 * {@code -1} if no match or {@code null} string input
 */
private static int ordinalIndexOf(final String str, final String searchStr, final int ordinal, final boolean lastIndex) {
 if (str == null || searchStr == null || ordinal &lt;= 0) {
 return -1;
 }
 if (searchStr.length() == 0) {
 return lastIndex ? str.length() : 0;
 }
 int found = 0;
 // set the initial index beyond the end of the string
 // this is to allow for the initial index decrement/increment
 int index = lastIndex ? str.length() : -1;
 do {
 if (lastIndex) {
 index = str.lastIndexOf(searchStr, index - 1); // step backwards thru string
 } else {
 index = str.indexOf(searchStr, index + 1); // step forwards through string
 }
 if (index &lt; 0) {
 return index;
 }
 found++;
 } while (found &lt; ordinal);
 return index;
}
</code></pre>

blocks|key|3230295|text|看起来你想要子串的字符串是一个文件路径。难道不能只按/拆分，然后从兴趣点开始使用数组条目吗？例如,|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3230296|String+folders+=+"/folder1/folder2/folder3/".split('/');
StringBuilder+subStringed+=+new+StringBuilder('/');
for+(int+i+=+2;+i+<+folders.length;+i%2B%2B)+{
++subStringed.append(folders[i]).append('/').;
}
System.out.println(subStringed.toString());|code-block|syntax|javascript|3230297|entityMap^0|Q|1|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|$]]

It looks like the string you want to substring is a file path. Can't you just split by <code>/</code> and then consume the array entries from the point of interest onward? For example,
<pre><code>String folders = &quot;/folder1/folder2/folder3/&quot;.split('/');
StringBuilder subStringed = new StringBuilder('/');
for (int i = 2; i &lt; folders.length; i++) {
 subStringed.append(folders[i]).append('/').;
}
System.out.println(subStringed.toString());

</code></pre>

blocks|key|294512|text|static+int+nthOccurrenceOfChar(String+str,+int+n,+char+ch)+{
++++int+count+=+0;
++++for+(int+i+=+0;+i+<+str.length();+i%2B%2B)
++++++++if+(str.charAt(i)+==+ch+&&+%2B%2Bcount+==+n)
++++++++++++return+i;
++++return+-1;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|294513|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>static int nthOccurrenceOfChar(String str, int n, char ch) {
 int count = 0;
 for (int i = 0; i &lt; str.length(); i++)
 if (str.charAt(i) == ch &amp;&amp; ++count == n)
 return i;
 return -1;
}
</code></pre>

Similar to a question posted <a href="https://stackoverflow.com/questions/2571716/find-nth-occurrence-of-a-character-in-a-string" title="Find Nth occurrence of a character in a string">here</a>, am looking
for a solution in Java.

That is, how to find the index of nth occurrence of a character/string from a string?

Example: "/folder1/folder2/folder3/". 
In this case, if I ask for 3rd occurrence of slash (/), it appears before folder3, and I expect to return this index position. My actual intention is to substring it from nth occurrence of a character.

Is there any convenient/ready-to-use method available in Java API or do we need to write a small logic on our own to solve this?

Also, 

<ol>
<li>I quickly searched whether any method is supported for this purpose at Apache Commons Lang's <a href="http://commons.apache.org/lang/api/org/apache/commons/lang3/StringUtils.html" rel="noreferrer" title="org.apache.commons.lang3.StringUtils">StringUtils</a>, but I don't find any.</li>
<li>Can regular expressions help in this regard?</li>
</ol>

How to find nth occurrence of character in a string?

Java

类似于上发布的一个问题，我正在寻找Java的解决方案。也就是说，如何从字符串中找到第n次出现的字符串的索引？示例："/folder1/folder2/folder3/".在本例中，如果我请求第三次出现斜杠(/)，它会出现在folder3之前，并且我希望返回这个索引位置。我的实际意图是从第n个出现的字符中子串它。Java...

问如何查找字符串中第n次出现的字符？
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