blocks|key|14297|text|使用重载版本的indexOf()，它将起始索引(fromIndex)作为第二个参数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|14298|str.indexOf("is",+str.indexOf("is")+%2B+1);|code-block|syntax|javascript|14299|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#indexOf-java.lang.String-int-^0|7|9|7|9|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use overloaded version of <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#indexOf-java.lang.String-int-" rel="noreferrer"><code>indexOf()</code></a>, which takes the starting index (fromIndex) as 2nd parameter:

<pre><code>str.indexOf("is", str.indexOf("is") + 1);
</code></pre>

blocks|key|14397|text|我正在使用：Apache+Commons+Lang:+StringUtils.ordinalIndexOf()|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|14398|StringUtils.ordinalIndexOf("Java+Language",+"a",+2)|code-block|syntax|javascript|14399|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#ordinalIndexOf(java.lang.CharSequence,%2520java.lang.CharSequence,%2520int)^0|6|1D|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]]]]

I am using: 
<a href="https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#ordinalIndexOf(java.lang.CharSequence,%20java.lang.CharSequence,%20int)" rel="noreferrer">Apache Commons Lang: StringUtils.ordinalIndexOf()</a>

<pre><code>StringUtils.ordinalIndexOf("Java Language", "a", 2)
</code></pre>

blocks|key|14267|text|int+first+=+string.indexOf("is");
int+second+=+string.indexOf("is",+first+%2B+1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|14268|这个重载开始从给定的索引中查找子字符串。|unstyled|14269|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>int first = string.indexOf("is");
int second = string.indexOf("is", first + 1);
</code></pre>

This overload starts looking for the substring from the given index.

blocks|key|230684|text|你可以写一个函数来返回出现位置的数组，Java有一个非常方便的String.regionMatches函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|230685|public+static+ArrayList<Integer>+occurrencesPos(String+str,+String+substr)+{
++++final+boolean+ignoreCase+=+true;
++++int+substrLength+=+substr.length();
++++int+strLength+=+str.length();

++++ArrayList<Integer>+occurrenceArr+=+new+ArrayList<Integer>();

++++for(int+i+=+0;+i+<+strLength+-+substrLength+%2B+1;+i%2B%2B)+{
++++++++if(str.regionMatches(ignoreCase,+i,+substr,+0,+substrLength))++{
++++++++++++occurrenceArr.add(i);
++++++++}
++++}
++++return+occurrenceArr;
}|code-block|syntax|javascript|230686|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 can write a function to return array of occurrence positions, Java has String.regionMatches function which is quite handy

<pre><code>public static ArrayList&lt;Integer&gt; occurrencesPos(String str, String substr) {
 final boolean ignoreCase = true;
 int substrLength = substr.length();
 int strLength = str.length();

 ArrayList&lt;Integer&gt; occurrenceArr = new ArrayList&lt;Integer&gt;();

 for(int i = 0; i &lt; strLength - substrLength + 1; i++) {
 if(str.regionMatches(ignoreCase, i, substr, 0, substrLength)) {
 occurrenceArr.add(i);
 }
 }
 return occurrenceArr;
}
</code></pre>

blocks|key|14482|text|我希望我来参加聚会没有迟到。这是我的答案。我喜欢使用Pattern/Matcher，因为它使用regex，它应该更高效。然而，我认为这个答案可以增强：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|14483|++++Matcher+matcher+=+Pattern.compile("is").matcher("I+think+there+is+a+smarter+solution,+isn't+there?");
++++int+numOfOcurrences+=+2;
++++for(int+i+=+0;+i+<+numOfOcurrences;+i%2B%2B)+matcher.find();
++++System.out.println("Index:+"+%2B+matcher.start());|code-block|syntax|javascript|14484|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 hope I'm not late to the party.. Here is my answer. I like using Pattern/Matcher because it uses regex which should be more efficient. Yet, I think this answer could be enhanced: 
<pre><code> Matcher matcher = Pattern.compile(&quot;is&quot;).matcher(&quot;I think there is a smarter solution, isn't there?&quot;);
 int numOfOcurrences = 2;
 for(int i = 0; i &lt; numOfOcurrences; i++) matcher.find();
 System.out.println(&quot;Index: &quot; + matcher.start());
</code></pre>

blocks|key|230643|text|我认为可以使用循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|230644|1+-+check+if+the+last+index+of+substring+is+not+the+end+of+the+main+string.
2+-+take+a+new+substring+from+the+last+index+of+the+substring+to+the+last+index+of+the+main+string+and+check+if+it+contains+the+search+string
3+-+repeat+the+steps+in+a+loop|code-block|syntax|javascript|230645|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 think a loop can be used.

<pre><code>1 - check if the last index of substring is not the end of the main string.
2 - take a new substring from the last index of the substring to the last index of the main string and check if it contains the search string
3 - repeat the steps in a loop
</code></pre>

blocks|key|230755|text|如果要查找两个以上匹配项的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|230756|public+static+int+ordinalIndexOf(String+fullText,String+subText,int+pos){

++++if(fullText.contains(subText)){
++++++++if(pos+<=+1){
++++++++++++return+fullText.indexOf(subText);
++++++++}else{
++++++++++++--pos;
++++++++++++return+fullText.indexOf(subText,+(+ordinalIndexOf(fullText,subText,pos)+%2B+1)+);
++++++++}
++++}else{
++++++++return+-1;
++++}

}|code-block|syntax|javascript|230757|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 want to find index for more than 2 occurrence:
<pre><code>public static int ordinalIndexOf(String fullText,String subText,int pos){

 if(fullText.contains(subText)){
 if(pos &lt;= 1){
 return fullText.indexOf(subText);
 }else{
 --pos;
 return fullText.indexOf(subText, ( ordinalIndexOf(fullText,subText,pos) + 1) );
 }
 }else{
 return -1;
 }

}
</code></pre>

blocks|key|14504|text|这看起来是个不错的聚会。我加入了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|14505|public+static+int+nthIndexOf(String+str,+String+subStr,+int+count)+{
++++int+ind+=+-1;
++++while(count+>+0)+{
++++++++ind+=+str.indexOf(subStr,+ind+%2B+1);
++++++++if(ind+==+-1)+return+-1;
++++++++count--;
++++}
++++return+ind;
}|code-block|syntax|javascript|14506|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|$]]

It seems to be a good party... I'm in:
<pre><code>public static int nthIndexOf(String str, String subStr, int count) {
 int ind = -1;
 while(count &gt; 0) {
 ind = str.indexOf(subStr, ind + 1);
 if(ind == -1) return -1;
 count--;
 }
 return ind;
}
</code></pre>

We are given a string, say, <code>"itiswhatitis"</code> and a substring, say, <code>"is"</code>.
I need to find the index of <code>'i'</code> when the string <code>"is"</code> occurs a second time in the original string.

<code>String.indexOf("is")</code> will return 2 in this case. I want the output to be 10 in this case.

Finding second occurrence of a substring in a string in Java

我们得到一个字符串，比方说"itiswhatitis"和一个子字符串，比方说"is"。当字符串"is"在原始字符串中第二次出现时，我需要找到'i'的索引。在这种情况下，String.indexOf("is")将返回2。在本例中，我希望输出为10。

问在Java中查找字符串中的子字符串的第二次出现
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回答 8

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