blocks|key|1003295|text|还有另一个版本的indexOf方法，以fromIndex作为参数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1003296|因此，您可以在循环中调用它，每次都将prevPosition+%2B+1作为第二个参数传递。|1003297|文档：|1003298|http://download.oracle.com/javase/6/docs/api/java/lang/String.html#indexOf(int,+int)|1003299|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/6/docs/api/java/lang/String.html#indexOf%2528int,%2520int%2529^0|8|7|J|9|0|I|G|0|0|0|2C|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|11|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|L|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|M|$N|$5|O|P|Q|E|$R|S]]]]

There's a another version of <code>indexOf</code> method, taking <code>fromIndex</code> as parameter. 
So, you can call it in a loop, each time passing <code>prevPosition + 1</code> as a second parameter.

Documentation: 
<a href="http://download.oracle.com/javase/6/docs/api/java/lang/String.html#indexOf%28int,%20int%29" rel="nofollow">http://download.oracle.com/javase/6/docs/api/java/lang/String.html#indexOf(int, int)</a>

blocks|key|1003317|text|您可以使用indexOf指定起始索引。因此，在您的循环中存储'x‘的最后一个位置，然后使用该索引%2B+1再次搜索。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1003318|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#indexOf%2528int,%2520int%2529^0|5|7|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

You can specify the start index with <a href="http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#indexOf%28int,%20int%29" rel="nofollow">indexOf</a>. So, in your loop you store the last position of 'x', then search again using that index + 1.

blocks|key|1003337|text|indexOf方法还有第二个变体，它接受起始索引作为参数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1003338|i+=+str.indexOf('x');
while(i+>=+0)+{
+++++System.out.println(i);
+++++i+=+str.indexOf('x',+i%2B1);
}|code-block|syntax|javascript|1003339|entityMap^0|0|7|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|$]]

There is a second variant of the <code>indexOf</code> method, which takes a start-index as a parameter.

<pre><code>i = str.indexOf('x');
while(i &gt;= 0) {
 System.out.println(i);
 i = str.indexOf('x', i+1);
}
</code></pre>

blocks|key|1003383|text|下面是Streams+API的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1003384|String+text+=+"Lets+search+for+E+?";

final+char+key+=+"e".charAt(0);
int[]+indices+=+IntStream.range(0,+text.length())
++++.filter(i+->+text.charAt(i)+==+key)
++++.toArray();|code-block|syntax|javascript|1003385|输出将如下所示：|1003386|//1
//6|1003387|如果要在结果中添加大写字母，则需要使用以下命令更改filter的参数闭包：|offset|length|style|CODE|1003388|.filter(i+->+Character.toLowerCase(text.charAt(i))+==+key)|1003389|然后，输出将如下所示：|1003390|//1
//6
//16|1003391|此外，如果您只想使用结果或将其打印出来用于快速测试，则可以使用forEach终端操作来完成此操作：|1003392|IntStream.range(0,+text.length())
++++.filter(i+->+text.charAt(i)+==+key)
++++.forEach(System.out::println);|1003393|entityMap^0|0|0|0|0|P|6|0|0|0|0|V|7|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|16|8|@$M|17|N|18|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1A|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1C|8|@$M|1D|N|1E|O|P]]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|10|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|11|$]]

Here is the solution with Streams API : 

<pre class="lang-java prettyprint-override"><code>String text = "Lets search for E ?";

final char key = "e".charAt(0);
int[] indices = IntStream.range(0, text.length())
 .filter(i -&gt; text.charAt(i) == key)
 .toArray();
</code></pre>

Output will be like this :

<pre class="lang-java prettyprint-override"><code>//1
//6
</code></pre>

If you want to add upper case letters into your result, you need to change the <code>filter</code>'s argument closure with :

<pre class="lang-java prettyprint-override"><code>.filter(i -&gt; Character.toLowerCase(text.charAt(i)) == key)
</code></pre>

Then the output will be like this :

<pre><code>//1
//6
//16
</code></pre>

Additionally if you just want to use the results or print it out for quick testing purposes you can do it with using <code>forEach</code> terminal operation: 

<pre class="lang-java prettyprint-override"><code>IntStream.range(0, text.length())
 .filter(i -&gt; text.charAt(i) == key)
 .forEach(System.out::println);
</code></pre>

blocks|key|1003397|text|String+name+=+"alameer+kaiser+aziz";
String+found+=+"a";
int+num+=+name.indexOf(found);
while+(num+>=0)+{
++++System.out.println(num);
++++num+=+name.indexOf(found,num%2B1);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1003398|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>String name = "alameer kaiser aziz";
String found = "a";
int num = name.indexOf(found);
while (num &gt;=0) {
 System.out.println(num);
 num = name.indexOf(found,num+1);
}
</code></pre>

I had a question about the indexOf method. I want to find multiple cases of "X" in a string.

Suppose my string is "x is x is x is x", I want to find x in all of its index positions.
But how do you do this for multiple cases? Is this even possible with indexOf?

I did int temp = str.indexOf('x');
It find the first x. I tried to do a for loop where i is initialized to length of string and this did not work since I kept finding the first x over and over.

<pre><code>for (int y = temp1; y &gt;= 0;y-- ) 
{
 int temp = str.indexOf('x');
 System.out.println(temp);
}
</code></pre>

But this does not work. Am I supposed to use regex? Because I don't really know how to use regex method.

Any help would be appreciated, thanks!

Java indexOf method for multiple matches in String

我有一个关于indexOf方法的问题。我想在一个字符串中找到多个"X“的案例。假设我的字符串是"x is x"，我想在它的所有索引位置中找到x。但是如何在多个情况下做到这一点呢？使用indexOf可以做到这一点吗？我做了int temp = str.indexOf(' x ')；它找到了第一个x。我试图做一个for循环...

问用于字符串中多个匹配的Java indexOf方法
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