与此 preg_replace 等效的 Java 是什么？
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What is the Java equivalent to this preg_replace?
提问by celsowm
<?php
$str = "word <a href=\"word\">word</word>word word";
$str = preg_replace("/word(?!([^<]+)?>)/i","repl",$str);
echo $str;
# repl <word word="word">repl</word>
?>
source: http://pureform.wordpress.com/2008/01/04/matching-a-word-characters-outside-of-html-tags/
来源：http: //pureform.wordpress.com/2008/01/04/matching-a-word-characters-outside-of-html-tags/
Unfortunality my project needs a semantic libs avaliable only for Java...
不幸的是，我的项目需要一个仅适用于 Java 的语义库...
// Thanks Celso
// 谢谢塞尔索
回答by kolrie
Use the String.replaceAll() method:
使用 String.replaceAll() 方法：
class Test {
public static void main(String[] args) {
String str = "word <a href=\"word\">word</word>word word";
str = str.replaceAll("word(?!([^<]+)?>)", "repl");
System.out.println(str);
}
}
Hope this helps.
希望这可以帮助。
回答by Alan Moore
To translate that regex for use in Java, all you have to do is get rid of the /delimiters and change the trailing ito an inline modifier, (?i). But it's not a very good regex; I would use this instead:
要将该正则表达式转换为在 Java 中使用，您所要做的就是去掉/分隔符并将尾随更改为i内联修饰符(?i). 但这不是一个很好的正则表达式；我会用这个代替：
(?i)word(?![^<>]++>)
According to RegexBuddy's Debug feature, when it tries to match the wordin <a href="word">, the original regex requires 23 steps to reject it, while this one takes only seven steps. The actual Java code is
根据 RegexBuddy 的 Debug 功能，当它尝试匹配wordin 时<a href="word">，原始正则表达式需要 23 步才能拒绝它，而这个只需 7 步。实际的Java代码是
str = str.replaceAll("(?i)word(?![^<>]++>)", "repl");
回答by Zak
Before providing a further answer, are you trying to parse an html document? If so, don't use regexes, use an html parser.
在提供进一步的答案之前，您是否尝试解析 html 文档？如果是这样，不要使用正则表达式，使用 html 解析器。
