blocks|key|5257110|text|substring()：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5257111|str.substring(startIndex,+endIndex);+|code-block|syntax|javascript|5257112|entityMap|0|LINK|mutability|MUTABLE|url|http://developer.android.com/reference/java/lang/String.html#substring%2528int%2529^0|0|B|0|B|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]]]]

<a href="http://developer.android.com/reference/java/lang/String.html#substring%28int%29"><code>substring()</code></a>: 

<pre><code>str.substring(startIndex, endIndex); 
</code></pre>

blocks|key|1730269|text|如果您知道开始和结束索引，则可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1730270|String+substr=mysourcestring.substring(startIndex,endIndex);|code-block|syntax|javascript|1730271|如果你想从特定的索引中得到子字符串，你可以使用：|1730272|String+substr=mysourcestring.substring(startIndex);|1730273|如果你想从头到尾得到特定字符的子串，你可以使用：|1730274|String+substr=mysourcestring.substring(mysourcestring.indexOf("characterValue"));|1730275|如果您想从特定字符后面获取子字符串，请将该数字添加到.indexOf(char)|offset|length|style|CODE|1730276|String+substr=mysourcestring.substring(mysourcestring.indexOf("characterValue")+%2B+1);|1730277|entityMap^0|0|0|0|0|0|0|Q|E|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

If you know the Start and End index, you can use 

<pre><code>String substr=mysourcestring.substring(startIndex,endIndex);
</code></pre>

If you want to get substring from specific index till end you can use :

<pre><code>String substr=mysourcestring.substring(startIndex);
</code></pre>

If you want to get substring from specific character till end you can use :

<pre><code>String substr=mysourcestring.substring(mysourcestring.indexOf("characterValue"));
</code></pre>

If you want to get substring from after a specific character, add that number to <code>.indexOf(char)</code>:

<pre><code>String substr=mysourcestring.substring(mysourcestring.indexOf("characterValue") + 1);
</code></pre>

blocks|key|1730282|text|使用android中的文本类：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1730283|TextUtils.substring+(charsequence+source,+int+start,++int+end)|offset|length|style|CODE|1730284|entityMap^0|0|0|1Q|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

use text untold class from android: <code>TextUtils.substring (charsequence source, int start, int end)</code>

blocks|key|1730291|text|您可以使用subSequence，它与C中的substr相同|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1730292|+Str.subSequence(int+Start+,+int+End)|code-block|syntax|javascript|1730293|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 use subSequence , it's same as substr in C

<pre><code> Str.subSequence(int Start , int End)
</code></pre>

blocks|key|5257171|text|下面是一个真实的例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5257172|String+hallostring+=+"hallo";
String+asubstring+=+hallostring.substring(0,+1);+|code-block|syntax|javascript|5257173|在示例中，子字符串将返回:h|offset|length|style|BOLD|5257174|entityMap^0|0|0|3|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here is a real world example:

<pre><code>String hallostring = "hallo";
String asubstring = hallostring.substring(0, 1); 
</code></pre>

In the example asubstring would return: h

blocks|key|1483406|text|在Android中获取子串的最好方法是使用(正如@user2503849所说的)+TextUtlis.substring(CharSequence,+int,+int)方法。我可以解释为什么。如果您将查看android.jar+(最新的API22)中的String.substring(int,+int)方法，您将看到：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1483407|public+String+substring(int+start)+{
++++if+(start+==+0)+{
++++++++return+this;
++++}
++++if+(start+>=+0+&&+start+<=+count)+{
++++++++return+new+String(offset+%2B+start,+count+-+start,+value);
++++}
++++throw+indexAndLength(start);
}|code-block|syntax|javascript|1483408|好的，然后..。你觉得私有构造函数String(int,+int,+char[])是什么样子的？|1483409|String(int+offset,+int+charCount,+char[]+chars)+{
++++this.value+=+chars;
++++this.offset+=+offset;
++++this.count+=+charCount;
}|1483410|正如我们所看到的，它保持了对“旧”值char[]数组的引用。因此，GC不能释放它。|1483411|在最新的Java中，它被修复了：|1483412|String(int+offset,+int+charCount,+char[]+chars)+{
++++this.value+=+Arrays.copyOfRange(chars,+offset,+offset+%2B+charCount);
++++this.offset+=+offset;
++++this.count+=+charCount;
}|1483413|Arrays.copyOfRange(...)在内部使用本机数组复制。|1483414|就是这样:)|1483415|诚挚的问候!|1483416|entityMap^0|15|17|2U|B|3I|Q|0|0|H|O|0|0|I|6|0|0|0|0|N|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1H|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|W|3|X|5|6|7|1M|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1N|8|@]|D|@]|E|$]]|$1|10|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|11|$]]

The best way to get substring in Android is using (as @user2503849 said) <code>TextUtlis.substring(CharSequence, int, int)</code> method. I can explain why. If you will take a look at the <code>String.substring(int, int)</code> method from <code>android.jar</code> (newest API 22), you will see:

<pre><code>public String substring(int start) {
 if (start == 0) {
 return this;
 }
 if (start &gt;= 0 &amp;&amp; start &lt;= count) {
 return new String(offset + start, count - start, value);
 }
 throw indexAndLength(start);
}
</code></pre>

Ok, than... How do you think the private constructor <code>String(int, int, char[])</code> looks like?

<pre><code>String(int offset, int charCount, char[] chars) {
 this.value = chars;
 this.offset = offset;
 this.count = charCount;
}
</code></pre>

As we can see it keeps reference to the "old" value <code>char[]</code> array. So, the GC can not free it.

In the newest Java it was fixed:

<pre><code>String(int offset, int charCount, char[] chars) {
 this.value = Arrays.copyOfRange(chars, offset, offset + charCount);
 this.offset = offset;
 this.count = charCount;
}
</code></pre>

<code>Arrays.copyOfRange(...)</code> uses native array copying inside.

That's it :)

Best regards!

blocks|key|1730374|text|您可以使用以下代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1730375|++++public+static+String+getSubString(String+mainString,+String+lastString,+String+startString)+{
++++String+endString+=+"";
++++int+endIndex+=+mainString.indexOf(lastString);
++++int+startIndex+=+mainString.indexOf(startString);
++++Log.d("message",+""+%2B+mainString.substring(startIndex,+endIndex));
++++endString+=+mainString.substring(startIndex,+endIndex);
++++return+endString;
}|code-block|syntax|javascript|1730376|在这个mainString中是一个超级string.like+"I_AmANDROID.Devloper“，而lastString是一个类似于”“的字符串。而startString就像“_”。所以这个function返回"AmANDROID“。享受您的代码时间。：)|offset|length|style|CODE|1730377|entityMap^0|0|0|3|A|1J|A|2S|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

you can use this code

<pre><code> public static String getSubString(String mainString, String lastString, String startString) {
 String endString = "";
 int endIndex = mainString.indexOf(lastString);
 int startIndex = mainString.indexOf(startString);
 Log.d("message", "" + mainString.substring(startIndex, endIndex));
 endString = mainString.substring(startIndex, endIndex);
 return endString;
}
</code></pre>

in this <code>mainString</code> is a Super string.like 
"I_AmANDROID.Devloper"
and <code>lastString</code> is a string like"." and startString is like"_".
so this <code>function</code> returns "AmANDROID".
enjoy your code time.:)

blocks|key|5257234|text|如果你想得到一个字符前后的子串，还有另一种方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5257235|String+s+="123dance456";
String[]+split+=+s.split("dance");
String+firstSubString+=+split[0];
String+secondSubString+=+split[1];|code-block|syntax|javascript|5257236|查看此帖子-+how+to+find+before+and+after+sub-string+in+a+string|offset|length|5257237|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/8871682/how-to-find-before-and-after-sub-string-in-a-string^0|0|0|7|1F|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

There is another way , if you want to get sub string before and after a character

<pre><code>String s ="123dance456";
String[] split = s.split("dance");
String firstSubString = split[0];
String secondSubString = split[1];
</code></pre>

check this post- 
<a href="https://stackoverflow.com/questions/8871682/how-to-find-before-and-after-sub-string-in-a-string">how to find before and after sub-string in a string</a>

blocks|key|5257256|text|substring(int+startIndex,+int+endIndex)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5257257|5257258|如果不指定endIndex，该方法将返回startIndex中的所有字符。|blockquote|5257259|5257260|startIndex+:起始索引是包含索引|5257261|endIndex+:结束索引是独占的|5257262|5257263|示例：|5257264|String+str+=+"abcdefgh"|5257265|str.substring(0,+4)+=>+abcd+abcd+abcd|5257266|str.substring(4,+6)+=>+ef|5257267|str.substring(6)+=>+gh|5257268|entityMap^0|0|13|0|0|0|0|0|0|0|0|0|N|0|0|J|0|0|J|0|0|G|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|14|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|15|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|16|8|@]|D|@]|E|$]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|18|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1A|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|T|3|U|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|1H|8|@$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|X|3|Y|5|6|7|1K|8|@$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|10|$]]

<code>substring(int startIndex, int endIndex)</code> 

<blockquote>
 If you don't specify endIndex, the method will return all the
 characters from startIndex.
 
 startIndex : starting index is inclusive
 
 endIndex : ending index is exclusive
</blockquote>

Example:

<code>String str = "abcdefgh"</code>

<code>str.substring(0, 4)</code> => abcd

<code>str.substring(4, 6)</code> => ef

<code>str.substring(6)</code> => gh

blocks|key|1487744|text|查找匹配模式的子字符串的多个匹配项时|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1487745|++++String+input_string+=+"foo/adsfasdf/adf/bar/erqwer/";
++++String+regex+=+"(foo/%7Cbar/)";+//+Matches+'foo/'+and+'bar/'

++++Pattern+pattern+=+Pattern.compile(regex);
++++Matcher+matcher+=+pattern.matcher(input_string);

++++while(matcher.find())+{
++++++String+str_matched+=+input_string.substring(matcher.start(),+matcher.end());
++++++++//+Do+something+with+a+match+found
++++}|code-block|syntax|javascript|1487746|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|$]]

When finding multiple occurrences of a substring matching a pattern 

<pre><code> String input_string = "foo/adsfasdf/adf/bar/erqwer/";
 String regex = "(foo/|bar/)"; // Matches 'foo/' and 'bar/'

 Pattern pattern = Pattern.compile(regex);
 Matcher matcher = pattern.matcher(input_string);

 while(matcher.find()) {
 String str_matched = input_string.substring(matcher.start(), matcher.end());
 // Do something with a match found
 }
</code></pre>

blocks|key|1730502|text|所有的应答者都给出了很好的答案。但是，我会给你所有相关的方法，这样任何人都可以从一个地方得到，如果我找到了新的东西，我会编辑我的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1730503|1)子字符串(0)-用于从给定的特定字符中剪切字符串。|1730504|2)子串(0，2)-给出起始(0)和结束(2)字符的子串。|1730505|3)+split+(+"NAME“)-将字符串分成两部分返回。第一部分是在split”NAME“中使用，另一部分是字符串的其余部分组合。|1730506|4)subSequence(0,3)+-返回给定开始(0)和结束索引(3)的序列。|1730507|*此字符串不是专门用于拆分字符串，但它可能对某些字符串使用完整|1730508|5)startswith("A",3)-返回特定起始字符的字符串。|1730509|例如,|1730510|String+s+=+"StackOverflow";
String[]+split+=+s.split("Stack");
++++++++++++System.out.println("STRING+NAME:"%2Bs.substring(2));
++++++++++++System.out.println("STRING+NAME:"%2Bs.substring(2,3));
++++++++++++System.out.println("STRING+NAME:"%2Bsplit[1]);
++++++++++++System.out.println("STRING+NAME:"%2Bsplit[0]);
++++++++++++System.out.println("STRING+NAME:"%2Bs.subSequence(2,5));|code-block|syntax|javascript|1730511|输出:+1)ackOverflow|1730512|2)a|1730513|3)溢出|1730514|4)堆栈|1730515|5)确认|1730516|我希望这将为您提供所需的足够信息。|1730517|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1A|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1B|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|1C|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1D|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1E|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1F|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1G|8|@]|9|@]|A|$S|T]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1J|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|1K|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|1L|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|1M|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|17|$]]

All of The responders gave good answers. However, I am giving you all relatable methods for this so that any one can get from one place, I'll edit my answer if I find something new.
1)substring(0)- use for cut string from given specific char.
2)Substring(0,2)- give you sub string from starting(0) and ending(2) characters.
3)Split(&quot;NAME&quot;)- return you string in two parts first is that you use in split &quot;NAME&quot; and another part is rest of string combine.
4)subSequence(0,3) - returns sequence of give start(0) and ending index(3).
*This one is not speciffically use for split string but though it may be use full for some one
5)startswith(&quot;A&quot;,3)- returns string for specific starting charecter.
for Example,
<pre><code>String s = &quot;StackOverflow&quot;;
String[] split = s.split(&quot;Stack&quot;);
 System.out.println(&quot;STRING NAME:&quot;+s.substring(2));
 System.out.println(&quot;STRING NAME:&quot;+s.substring(2,3));
 System.out.println(&quot;STRING NAME:&quot;+split[1]);
 System.out.println(&quot;STRING NAME:&quot;+split[0]);
 System.out.println(&quot;STRING NAME:&quot;+s.subSequence(2,5));
</code></pre>
Output:
1)ackOverflow
2)a
3)Overflow
4)stack
5)ack
I hope this will give you enough information that you require.

what is the best way to extract a substring from a string in android?

Extract substring from a string

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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在android中，从字符串中提取子字符串的最佳方法是什么？

问从字符串中提取子字符串
EN

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问从字符串中提取子字符串EN