entityMap|blocks|key|7g0tr|text|你可以写一个循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|8tc1j|public+static+void+replaceAll(StringBuilder+builder,+String+from,+String+to)+{
++++int+index+=+builder.indexOf(from);
++++while+(index+!=+-1)+{
++++++++builder.replace(index,+index+%2B+from.length(),+to);
++++++++index+%2B=+to.length();+//+Move+to+the+end+of+the+replacement
++++++++index+=+builder.indexOf(from,+index);
++++}
}|code-block|syntax|javascript|7dfqt|请注意，在某些情况下，使用它可能会更快|3e0i3|，从后面工作。我怀疑如果你用一个短字符串替换一个长字符串，情况就是这样--所以当你开始的时候，任何替换字符串都有更少的需要复制的东西。无论如何，这应该给你一个起点。^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|M|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|N|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|O|9|@]|A|@]|B|$]]]]

Well, you can write a loop:
<pre><code>public static void replaceAll(StringBuilder builder, String from, String to) {
 int index = builder.indexOf(from);
 while (index != -1) {
 builder.replace(index, index + from.length(), to);
 index += to.length(); // Move to the end of the replacement
 index = builder.indexOf(from, index);
 }
}
</code></pre>
Note that in some cases it may be faster to use <code>lastIndexOf</code>, working from the back. I suspect that's the case if you're replacing a long string with a short one - so when you get to the start, any replacements have less to copy. Anyway, this should give you a starting point.

entityMap|blocks|key|8274n|text|你可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|fq5kc|模式|dv02n|/|7npk5|匹配器|c87up|..。在Matcher+javadocs中：|23tvu|Pattern+p+=+Pattern.compile("cat");
+Matcher+m+=+p.matcher("one+cat+two+cats+in+the+yard");
+StringBuffer+sb+=+new+StringBuffer();
+while+(m.find())+{
+++++m.appendReplacement(sb,+"dog");
+}
+m.appendTail(sb);
+System.out.println(sb.toString());|code-block|syntax|javascript^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|Q|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|R|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|S|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|T|9|@]|A|@]|B|$]]|$2|K|4|L|6|M|8|U|9|@]|A|@]|B|$N|O]]]]

You could use <a href="http://download.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html" rel="noreferrer">Pattern</a>/<a href="http://download.oracle.com/javase/6/docs/api/java/util/regex/Matcher.html" rel="noreferrer">Matcher</a>. From the Matcher javadocs:

<pre><code> Pattern p = Pattern.compile("cat");
 Matcher m = p.matcher("one cat two cats in the yard");
 StringBuffer sb = new StringBuffer();
 while (m.find()) {
 m.appendReplacement(sb, "dog");
 }
 m.appendTail(sb);
 System.out.println(sb.toString());
</code></pre>

entityMap|blocks|key|fe1mr|text|@Adam:我认为在你的代码片段中，你应该跟踪m.find()的起始位置，因为字符串替换可能会改变最后一个字符匹配后的偏移量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|24qnd|public+static+void+replaceAll(StringBuilder+sb,+Pattern+pattern,+String+replacement)+{
++++Matcher+m+=+pattern.matcher(sb);
++++int+start+=+0;
++++while+(m.find(start))+{
++++++++sb.replace(m.start(),+m.end(),+replacement);
++++++++start+=+m.start()+%2B+replacement.length();
++++}
}|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

@Adam: I think in your code snippet you should track the start position for m.find() because string replacement may change the offset after the last character matched.

<pre><code>public static void replaceAll(StringBuilder sb, Pattern pattern, String replacement) {
 Matcher m = pattern.matcher(sb);
 int start = 0;
 while (m.find(start)) {
 sb.replace(m.start(), m.end(), replacement);
 start = m.start() + replacement.length();
 }
}
</code></pre>

entityMap|blocks|key|dh0j1|text|看看JavaDoc+of|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7l6lq|replaceAll|7q2ks|方法|9t05q|字符串|8nim4|类：|ohgc|将此字符串中与给定正则表达式匹配的每个子字符串替换为给定的替换。调用此str.replaceAll(regex，repl)形式的方法会产生与表达式完全相同的结果|1frmv|6oavg|java.util.regex.Pattern.compile(regex).matcher(str).replaceAll(repl)|68uon|8kfma|4c41r|如您所见，您可以使用|3e1p3|模式|flim6|和|1qusn|匹配器|1iuv9|要做到这一点。^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|11|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|12|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|13|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|14|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|15|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|16|9|@]|A|@]|B|$]]|$2|M|4|-4|6|7|8|17|9|@]|A|@]|B|$]]|$2|N|4|O|6|7|8|18|9|@]|A|@]|B|$]]|$2|P|4|-4|6|7|8|19|9|@]|A|@]|B|$]]|$2|Q|4|-4|6|7|8|1A|9|@]|A|@]|B|$]]|$2|R|4|S|6|7|8|1B|9|@]|A|@]|B|$]]|$2|T|4|U|6|7|8|1C|9|@]|A|@]|B|$]]|$2|V|4|W|6|7|8|1D|9|@]|A|@]|B|$]]|$2|X|4|Y|6|7|8|1E|9|@]|A|@]|B|$]]|$2|Z|4|10|6|7|8|1F|9|@]|A|@]|B|$]]]]

Look at JavaDoc of replaceAll method of String class:

<blockquote>
 Replaces each substring of this string that matches the given regular
 expression with the given replacement. An invocation of this method
 of the form str.replaceAll(regex, repl) yields exactly the same result
 as the expression 
 
 java.util.regex.Pattern.compile(regex).matcher(str).replaceAll(repl)
</blockquote>

As you can see you can use Pattern and Matcher to do that.

entityMap|blocks|key|bli4j|text|这个类|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1n7d1|在|ae2aa|Apache+Commons+Lang|6vjj7|允许替换：|ddqpe|public+StrBuilder+replaceAll(String+searchStr,+String+replaceStr)|code-block|syntax|javascript|2qk57|*|85b0j|8k02h|它接收的不是一个正则表达式，而是一个简单的字符串。^0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|T|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|U|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|V|9|@]|A|@]|B|$]]|$2|I|4|J|6|K|8|W|9|@]|A|@]|B|$L|M]]|$2|N|4|O|6|7|8|X|9|@]|A|@]|B|$]]|$2|P|4|-4|6|7|8|Y|9|@]|A|@]|B|$]]|$2|Q|4|R|6|7|8|Z|9|@]|A|@]|B|$]]]]

The class <a href="https://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/text/StrBuilder.html" rel="noreferrer"><code>org.apache.commons.lang3.text.StrBuilder</code></a> in <a href="http://Apache%20Commons%20Lang" rel="noreferrer">Apache Commons Lang</a> allows replacements:

<pre><code>public StrBuilder replaceAll(String searchStr, String replaceStr)
</code></pre>

* This does not receive a regular expression but a simple string.

entityMap|blocks|key|c0536|text|即使是简单的方法也是使用字符串ReplaceAll函数本身。你可以把它写成|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1lud8|StringBuilder+sb+=+new+StringBuilder("Hi+there,+are+you+there?")
System.out.println(Pattern.compile("there").matcher(sb).replaceAll("niru"));|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Even simple one is using the String ReplaceAll function itself. You can write it as

<pre><code>StringBuilder sb = new StringBuilder("Hi there, are you there?")
System.out.println(Pattern.compile("there").matcher(sb).replaceAll("niru"));
</code></pre>

entityMap|blocks|key|1qrq7|text|是的。它使用起来非常简单|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500m4|方法：|6k61h|package+com.test;

public+class+Replace+{

++++public+static+void+main(String[]+args)+{
++++++++String+input+=+"Hello+World";
++++++++input+=+input.replaceAll("o",+"0");
++++++++System.out.println(input);
++++}
}|code-block|syntax|javascript|1p403|输出：|9hnj6|Hell0+W0rld|aeq83|如果您真的想使用|73ql8|取而代之的是，你可以这样做：|6rtp|public+static+void+main(String+args[])+{
++++StringBuilder+sb+=+new+StringBuilder("This+is+a+new+StringBuilder");

++++System.out.println("Before:+"+%2B+sb);

++++String+from+=+"new";
++++String+to+=+"replaced";
++++sb+=+sb.replace(sb.indexOf(from),+sb.indexOf(from)+%2B+from.length(),+to);

++++System.out.println("After:+"+%2B+sb);
}|d78rh|b6ol7|Before:+This+is+a+new+StringBuilder
After:+This+is+a+replaced+StringBuilder^0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|W|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|X|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|Y|9|@]|A|@]|B|$H|I]]|$2|J|4|K|6|7|8|Z|9|@]|A|@]|B|$]]|$2|L|4|M|6|G|8|10|9|@]|A|@]|B|$H|I]]|$2|N|4|O|6|7|8|11|9|@]|A|@]|B|$]]|$2|P|4|Q|6|7|8|12|9|@]|A|@]|B|$]]|$2|R|4|S|6|G|8|13|9|@]|A|@]|B|$H|I]]|$2|T|4|K|6|7|8|14|9|@]|A|@]|B|$]]|$2|U|4|V|6|G|8|15|9|@]|A|@]|B|$H|I]]]]

Yes. It is very simple to use <code>String.replaceAll()</code> method:

<pre><code>package com.test;

public class Replace {

 public static void main(String[] args) {
 String input = "Hello World";
 input = input.replaceAll("o", "0");
 System.out.println(input);
 }
}
</code></pre>

Output:

<pre><code>Hell0 W0rld
</code></pre>

If you really want to use <code>StringBuilder.replace(int start, int end, String str)</code> instead then here you go:

<pre><code>public static void main(String args[]) {
 StringBuilder sb = new StringBuilder("This is a new StringBuilder");

 System.out.println("Before: " + sb);

 String from = "new";
 String to = "replaced";
 sb = sb.replace(sb.indexOf(from), sb.indexOf(from) + from.length(), to);

 System.out.println("After: " + sb);
}
</code></pre>

Output:

<pre><code>Before: This is a new StringBuilder
After: This is a replaced StringBuilder
</code></pre>

entityMap|blocks|key|dt6mp|text|java.util.regex.Pattern.matcher(CharSequence+%25+s)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|dfe0d|可以使用StringBuilder作为参数，这样就可以在不调用builder.toString()的情况下使用start()和end()查找和替换出现的每个模式^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|E|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|F|9|@]|A|@]|B|$]]]]

<a href="http://download.oracle.com/javase/1.4.2/docs/api/java/util/regex/Pattern.html#matcher%28java.lang.CharSequence%29" rel="nofollow noreferrer">java.util.regex.Pattern.matcher(CharSequence s)</a> can use a StringBuilder as an argument so you can find and replace each occurence of your pattern using start() and end() without calling builder.toString()

entityMap|blocks|key|6p4ba|text|使用以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|92m84|/**
*+Utility+method+to+replace+the+string+from+StringBuilder.
*+@param+sb++++++++++the+StringBuilder+object.
*+@param+toReplace+++the+String+that+should+be+replaced.
*+@param+replacement+the+String+that+has+to+be+replaced+by.
*+
*/
public+static+void+replaceString(StringBuilder+sb,
+++++++++++++++++++++++++++++++++String+toReplace,
+++++++++++++++++++++++++++++++++String+replacement)+{++++++
++++int+index+=+-1;
++++while+((index+=+sb.lastIndexOf(toReplace))+!=+-1)+{
++++++++sb.replace(index,+index+%2B+toReplace.length(),+replacement);
++++}
}|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Use the following:

<pre><code>/**
* Utility method to replace the string from StringBuilder.
* @param sb the StringBuilder object.
* @param toReplace the String that should be replaced.
* @param replacement the String that has to be replaced by.
* 
*/
public static void replaceString(StringBuilder sb,
 String toReplace,
 String replacement) { 
 int index = -1;
 while ((index = sb.lastIndexOf(toReplace)) != -1) {
 sb.replace(index, index + toReplace.length(), replacement);
 }
}
</code></pre>

entityMap|blocks|key|3vrvg|text|这是一个将修改传入的StringBuilder的就地replaceAll。我想我应该发布这篇文章，因为我希望在不创建新字符串的情况下做replaceAll。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|nr7d|public+static+void+replaceAll(StringBuilder+sb,+Pattern+pattern,+String+replacement)+{
++++Matcher+m+=+pattern.matcher(sb);
++++while(m.find())+{
++++++++sb.replace(m.start(),+m.end(),+replacement);
++++}
}|code-block|syntax|javascript|49atp|我很震惊这样做的代码是如此简单(出于某种原因，我以为在使用匹配器时更改StringBuilder会抛出组开始/结束，但事实并非如此)。|789n7|这可能比其他正则表达式的答案更快，因为模式已经编译，您没有创建新的字符串，但我没有做任何基准测试。^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|M|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|N|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|O|9|@]|A|@]|B|$]]]]

Here is an in place replaceAll that will modify the passed in StringBuilder. I thought that I would post this as I was looking to do replaceAll with out creating a new String.

<pre><code>public static void replaceAll(StringBuilder sb, Pattern pattern, String replacement) {
 Matcher m = pattern.matcher(sb);
 while(m.find()) {
 sb.replace(m.start(), m.end(), replacement);
 }
}
</code></pre>

I was shocked how simple the code to do this was (for some reason I thought changing the StringBuilder while using the matcher would throw of the group start/end but it does not).

This is probably faster than the other regex answers because the pattern is already compiled and your not creating a new String but I didn't do any benchmarking.

entityMap|blocks|key|al231|text|public+static+String+replaceCharsNew(String+replaceStr,Map+replaceStrMap){
++++++++StringBuilder+replaceStrBuilder+=+new+StringBuilder(replaceStr);
++++++++Set+keys=replaceStrMap.keySet();
++++++++for(String+invalidChar:keys){
++++++++++++int+index+=+-1;
++++++++++++while((index=replaceStrBuilder.indexOf(invalidChar,index))+!=-1){
++++++++++++++++replaceStrBuilder.replace(index,index%2BinvalidChar.length(),replaceStrMap.get(invalidChar));
++++++++++++}
++++++++}
++++++++return+replaceStrBuilder.toString();
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript^0^^$0|$]|1|@$2|3|4|5|6|7|8|E|9|@]|A|@]|B|$C|D]]]]

<pre><code>public static String replaceCharsNew(String replaceStr,Map&lt;String,String&gt; replaceStrMap){
 StringBuilder replaceStrBuilder = new StringBuilder(replaceStr);
 Set&lt;String&gt; keys=replaceStrMap.keySet();
 for(String invalidChar:keys){
 int index = -1;
 while((index=replaceStrBuilder.indexOf(invalidChar,index)) !=-1){
 replaceStrBuilder.replace(index,index+invalidChar.length(),replaceStrMap.get(invalidChar));
 }
 }
 return replaceStrBuilder.toString();
 }
</code></pre>

entityMap|blocks|key|1prhf|text|创建一个方法并让|type|unstyled|depth|inlineStyleRanges|entityRanges|data|acm9o|为您做到这一点：|3tev0|public+static+void+replaceAll(StringBuilder+sb,+String+regex,+String+replacement)
{
++++String+aux+=+sb.toString();
++++aux+=+aux.replaceAll(regex,+replacement);
++++sb.setLength(0);
++++sb.append(aux);+++++
}|code-block|syntax|javascript^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|K|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|L|9|@]|A|@]|B|$H|I]]]]

How about create a method and let <code>String.replaceAll</code> do it for you:

<pre><code>public static void replaceAll(StringBuilder sb, String regex, String replacement)
{
 String aux = sb.toString();
 aux = aux.replaceAll(regex, replacement);
 sb.setLength(0);
 sb.append(aux); 
}
</code></pre>

entityMap|blocks|key|e41ij|text|我发现了这个方法:Matcher.replaceAll(字符串替换)；在java.util.regex.Matcher.java中可以看到更多：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6520h|/**
+*+Replaces+every+subsequence+of+the+input+sequence+that+matches+the
+*+pattern+with+the+given+replacement+string.
+*
+*++This+method+first+resets+this+matcher.++It+then+scans+the+input
+*+sequence+looking+for+matches+of+the+pattern.++Characters+that+are+not
+*+part+of+any+match+are+appended+directly+to+the+result+string;+each+match
+*+is+replaced+in+the+result+by+the+replacement+string.++The+replacement
+*+string+may+contain+references+to+captured+subsequences+as+in+the+{@link
+*+#appendReplacement+appendReplacement}+method.
+*
+*++Note+that+backslashes+(\)+and+dollar+signs+($)+in
+*+the+replacement+string+may+cause+the+results+to+be+different+than+if+it
+*+were+being+treated+as+a+literal+replacement+string.+Dollar+signs+may+be
+*+treated+as+references+to+captured+subsequences+as+described+above,+and
+*+backslashes+are+used+to+escape+literal+characters+in+the+replacement
+*+string.
+*
+*++Given+the+regular+expression+a*b,+the+input
+*+"aabfooaabfooabfoob",+and+the+replacement+string
+*+"-",+an+invocation+of+this+method+on+a+matcher+for+that
+*+expression+would+yield+the+string+"-foo-foo-foo-".
+*
+*++Invoking+this+method+changes+this+matcher's+state.++If+the+matcher
+*+is+to+be+used+in+further+matching+operations+then+it+should+first+be
+*+reset.++
+*
+*+@param++replacement
+*+++++++++The+replacement+string
+*
+*+@return++The+string+constructed+by+replacing+each+matching+subsequence
+*++++++++++by+the+replacement+string,+substituting+captured+subsequences
+*++++++++++as+needed
+*/
public+String+replaceAll(String+replacement)+{
++++reset();
++++StringBuffer+buffer+=+new+StringBuffer(input.length());
++++while+(find())+{
++++++++appendReplacement(buffer,+replacement);
++++}
++++return+appendTail(buffer).toString();
}|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

I found this method: Matcher.replaceAll(String replacement);
In java.util.regex.Matcher.java you can see more:

<pre><code> /**
 * Replaces every subsequence of the input sequence that matches the
 * pattern with the given replacement string.
 *
 * &lt;p&gt; This method first resets this matcher. It then scans the input
 * sequence looking for matches of the pattern. Characters that are not
 * part of any match are appended directly to the result string; each match
 * is replaced in the result by the replacement string. The replacement
 * string may contain references to captured subsequences as in the {@link
 * #appendReplacement appendReplacement} method.
 *
 * &lt;p&gt; Note that backslashes (&lt;tt&gt;\&lt;/tt&gt;) and dollar signs (&lt;tt&gt;$&lt;/tt&gt;) in
 * the replacement string may cause the results to be different than if it
 * were being treated as a literal replacement string. Dollar signs may be
 * treated as references to captured subsequences as described above, and
 * backslashes are used to escape literal characters in the replacement
 * string.
 *
 * &lt;p&gt; Given the regular expression &lt;tt&gt;a*b&lt;/tt&gt;, the input
 * &lt;tt&gt;"aabfooaabfooabfoob"&lt;/tt&gt;, and the replacement string
 * &lt;tt&gt;"-"&lt;/tt&gt;, an invocation of this method on a matcher for that
 * expression would yield the string &lt;tt&gt;"-foo-foo-foo-"&lt;/tt&gt;.
 *
 * &lt;p&gt; Invoking this method changes this matcher's state. If the matcher
 * is to be used in further matching operations then it should first be
 * reset. &lt;/p&gt;
 *
 * @param replacement
 * The replacement string
 *
 * @return The string constructed by replacing each matching subsequence
 * by the replacement string, substituting captured subsequences
 * as needed
 */
public String replaceAll(String replacement) {
 reset();
 StringBuffer buffer = new StringBuffer(input.length());
 while (find()) {
 appendReplacement(buffer, replacement);
 }
 return appendTail(buffer).toString();
}
</code></pre>

Am I missing something, or does StringBuilder lack the same "replace all occurrences of a string A with string B" function that the normal String class does? The StringBuilder replace function isn't quite the same. Is there any way to this more efficiently without generating multiple Strings using the normal String class?

Replace all occurrences of a String using StringBuilder?

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。

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 我是否遗漏了什么，或者StringBuilder缺少与普通string类相同的“用string B替换字符串A的所有匹配项”功能？StringBuilder的替换功能并不完全相同。有没有什么方法可以在不使用普通的String类生成多个String的情况下更有效地做到这一点呢？

问是否使用StringBuilder替换所有出现的字符串？
EN

回答 13

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问是否使用StringBuilder替换所有出现的字符串？EN

回答 13

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问是否使用StringBuilder替换所有出现的字符串？
EN