Java重复模式匹配 [英] Java repetitive pattern matching

问题描述

 （\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ [[^ \\ [] * \\]）* 
  

匹配任何字符串包含在[]中，只要它不包含[character]。例如，它将匹配

  [a] [nice] [repetitive] [pattern] 
  

没有现有的知识有多少这样的组存在，我找不到一种通过模式匹配器访问各个匹配组的方式，即无法获得

  [a]，[nice]，[repetitive]，[pattern] 
  

（或者更好的是没有括号的文本），在4个不同的字符串中。

使用pattern.matcher（）我总是得到最后一个组。

肯定有一个简单的方式在Java中这样做，我错过了

感谢任何帮助。

解决方案

 code> String string =[a] [nice] [repetitive] [pattern]; 
 String regexp =\\ [（[^ \\ [] *）\\]; 
模式模式= Pattern.compile（regexp）; 
 Matcher matcher = pattern.matcher（string）; 
 while（matcher.find（））{
 System.out.println（matcher.group（1））; 
} 
  

I am trying to get each of the repetitive matches of a simple regular expression in Java:

(\\[[^\\[]*\\])*

which matches any string enclosed in [], as long as it does not contain the [ character. For example, it would match

[a][nice][repetitive][pattern]

There is no prior knowledge of how many such groups exist and I cannot find a way of accessing the individual matching groups via a pattern matcher, i.e. can't get

[a], [nice], [repetitive], [pattern]

(or, even better, the text without the brackets), in 4 different strings.

Using pattern.matcher() I always get the last group.

Surely there must be a simple way of doing this in Java, which I am missing?

Thanks for any help.

解决方案

    String string = "[a][nice][repetitive][pattern]";
    String regexp = "\\[([^\\[]*)\\]";
    Pattern pattern = Pattern.compile(regexp);
    Matcher matcher = pattern.matcher(string);
    while (matcher.find()) {
        System.out.println(matcher.group(1));
    }

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