正则表达式找到所有匹配

3 回答

• 1

  您需要使用non-greedy匹配：

  start.*?end
  

  在您的情况下，模式是：

  !style_delete\\s\\[(.*?)\\] (Even simple to understand than first version :))
  

  证明（Java 7）：

  String string = "someother text !style_delete [company code : 43ev4] between text !style_delete [organiztion : 0asj9] end of line text"; 
  Pattern pattern = Pattern.compile("!style_delete\\s\\[(.*?)\\]");
  Matcher matcher = pattern.matcher(string) ;
  while (matcher.find()) {
      System.out.println(matcher.group());
  }
  

  链接到证据：http://ideone.com/Qtymb3

  回复于 2024-04-17T01:11:52+08:00
• 3

  @Test
  public void test() {
      final String input = "someother text !style_delete [company code : 43ev4] between text !style_delete [organiztion : 0asj9] end of line text";
      // my regexp:strong text
      // final String regex = "(!style_delete\\s\\[[a-zA-Z0-9\\s:]*\\])";
      // regexp from Trinmon:
      final String regex = "(!style_delete\\s*\\[[^\\]]*\\])";
  
      final Matcher m = Pattern.compile(regex).matcher(input);
  
      final List<String> matches = new ArrayList<>();
      while (m.find()) {
          matches.add(m.group(0));
      }
  
      assertEquals(2, matches.size());
      assertEquals("match 1: ", matches.get(0), "!style_delete [company code : 43ev4]");
      assertEquals("match 2: ", matches.get(1), "!style_delete [organiztion : 0asj9]");
  }
  

  edit

  也许Trinimon的答案模式更优雅一点 . 我用正则表达式更新了正则表达式 .

  回复于 2024-04-17T01:11:52+08:00
• 3

  这是因为 .* 贪婪 . 改为使用它：

  "!style_delete\\s*\\[[^\\]]*\\]"
  

  这意味着：匹配括号中的所有内容，不包括结束 ] .

  或者使 [] 之间的内容非贪婪：

  "!style_delete\\s*\\[.*?\\]"
  

  回复于 2024-04-17T01:11:52+08:00