blocks|key|1068762|text|字符串在Java语言中是不可变的，因此您需要构建一个新的java.util.String实例。为此，您必须解析原始字符串(例如，使用正则表达式)。StringBuilder是一个很好的工具。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1068763|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/StringBuilder.html^0|S|G|21|D|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<code>java.util.String</code>s are immutable in Java, so you need to build a new String instance. You have to parse the original string (e.g. using regexes) for that. <a href="http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/StringBuilder.html" rel="nofollow">StringBuilder</a> is a good utility for that.

blocks|key|1068793|text|为此，我将使用Matcher：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1068794|StringBuffer+sb+=+new+StringBuffer();
Matcher+m+=+Pattern.compile("\\d%2B").matcher(input);
while(m.find())+{
++++m.appendReplacement(sb,"$0");
}
m.appendTail(sb);
String+newString+=+sb.toString();|code-block|syntax|javascript|1068795|entityMap^0|7|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|$]]

I would use a <code>Matcher</code> for that:

<pre><code>StringBuffer sb = new StringBuffer();
Matcher m = Pattern.compile("\\d+").matcher(input);
while(m.find()) {
 m.appendReplacement(sb,"&lt;sub&gt;$0&lt;/sub&gt;");
}
m.appendTail(sb);
String newString = sb.toString();
</code></pre>

blocks|key|5106847|text|我不是java专家，但我很确定，你可以使用正则表达式替换函数，其中的模式是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5106848|(\d%2B)|code-block|syntax|javascript|5106849|替换字符串将为：|5106850|$1|5106851|抱歉，我不是100%25确定java语法，但它可能是这样的：|5106852|str+=+str.replaceAll("(\\d%2B)",+"$1");|5106853|此方法是在Java+5中添加的|5106854|以下是有关java中regexp的更多信息：|5106855|http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html|offset|length|5106856|http://www.regular-expressions.info/java.html|5106857|http://javamex.com/tutorials/regular_expressions/search_replace.shtml|5106858|entityMap|0|LINK|mutability|MUTABLE|url|1|2^0|0|0|0|0|0|0|0|0|0|1X|0|0|0|19|1|0|0|1X|2|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1B|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1D|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|1E|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1F|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1G|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1H|8|@]|9|@$U|1I|V|1J|1|1K]]|A|$]]|$1|W|3|X|5|6|7|1L|8|@]|9|@$U|1M|V|1N|1|1O]]|A|$]]|$1|Y|3|Z|5|6|7|1P|8|@]|9|@$U|1Q|V|1R|1|1S]]|A|$]]|$1|10|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|T]]|17|$5|13|14|15|A|$16|X]]|18|$5|13|14|15|A|$16|Z]]]]

I'm not a java expert, but I'm pretty sure, you can use regular expressions replace function, where the pattern will be:

<pre><code>(\d+)
</code></pre>

and replacement string will be:

<pre><code>&lt;sub&gt;$1&lt;/sub&gt;
</code></pre>

Sorry I'm not 100% sure about the java syntax, but it can be something like:

<pre><code>str = str.replaceAll("(\\d+)", "&lt;sub&gt;$1&lt;/sub&gt;");
</code></pre>

This method was added in Java 5

here are some more information about regexp in java:

<a href="http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html" rel="nofollow">http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html</a>

<a href="http://www.regular-expressions.info/java.html" rel="nofollow">http://www.regular-expressions.info/java.html</a>

<a href="http://javamex.com/tutorials/regular_expressions/search_replace.shtml" rel="nofollow">http://javamex.com/tutorials/regular_expressions/search_replace.shtml</a>

blocks|key|1068847|text|使用匹配器和Pattern类搜索pattern："\\d%2B"|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1068848|之后，找到起始和结束索引，并添加和'‘|1068849|entityMap^0|O|6|0|G|5|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

Use Matcher and Pattern classes with searching for pattern:
<code>"\\d+"</code>

After that, find starting and ending indexes and add <code>&lt;sub&gt;</code> and ''

blocks|key|1068903|text|为此，您需要构造一个新的String对象--没有办法“就地”更改String+(String是不可变的)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1068904|String.replaceAll将为您完成此操作。|offset|length|style|CODE|1068905|str+=+str.replaceAll(+"\\d%2B",+"$0"+);|code-block|syntax|javascript|1068906|就是你所需要的|1068907|entityMap^0|0|0|H|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You will need to construct a new String object for this - there is no way to change a String "in place" (Strings are immutable)

<code>String.replaceAll</code> will do this for you.

<pre><code>str = str.replaceAll( "\\d+", "&lt;sub&gt;$0&lt;/sub&gt;" );
</code></pre>

is all you need

blocks|key|5107037|text|字符串是不可变的，所以我们不能在现有的字符串中插入子标签。因此，您需要一个创建新字符串的方法。您可以使用regex或使用字符串构建器生成如下所示的新字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5107038|public+String+parseString(String+input)+{
++++++++StringBuilder+builder+=+new+StringBuilder();
++++++++String+format+=+"%25c";
++++++++for(char+c:input.toCharArray())+{
++++++++++++if+(Character.isDigit(c))+{
++++++++++++++++builder.append(String.format(format,+c));
++++++++++++}+else+{
++++++++++++++++builder.append(c);
++++++++++++}
++++++++}
++++++++return+builder.toString();
++++}|code-block|syntax|javascript|5107039|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|$]]

String are immutable, so we cant insert the sub tags in the existing string. So you would need a method that creates a new string. You could use regex, or use a string builder to generate a new string like this: 

<pre><code>public String parseString(String input) {
 StringBuilder builder = new StringBuilder();
 String format = "&lt;sub&gt;%c&lt;/sub&gt;";
 for(char c:input.toCharArray()) {
 if (Character.isDigit(c)) {
 builder.append(String.format(format, c));
 } else {
 builder.append(c);
 }
 }
 return builder.toString();
 }
</code></pre>

I have a String:

<pre><code>C6H14
</code></pre>

What I want it to be is:

<pre><code>C&lt;sub&gt;6&lt;/sub&gt;H&lt;sub&gt;14&lt;/sub&gt;
</code></pre>

So basically I want to add <code>&lt;sub&gt;</code> before a number and <code>&lt;/sub&gt;</code> after a number. How can I accomplish this? Keep in mind that the String can be indefinitely long:

<pre><code>X4Y6Z10W4T12
</code></pre>

Manipulate String by inserting at index

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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我有一个字符串：C6H14我想要的是：C6H14所以基本上我想在数字前面加上，在数字后面加上。我如何才能做到这一点呢？请记住，字符串可以无限长：X4Y6Z10W4T12

问通过插入at索引来操作字符串
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