Java正則表達式捕獲重復的組

Question

我無法通過正則表達式捕獲第2組中的所有數據，包括所有字符，數字，空格和特殊字符

嘗試過正則表達式

 final String regex = "^:(.*?)//(.*[\\s\\S]?)";
    String line1 = ":Humpty Dumpty sat on a wall";
    String line2 = "//Humpty Dumpty had a great fall";
    String rhyme = line1 +  line2+"\n"+ "ssdsds"+"\n";
    final String value = rhyme.replaceAll(regex , "$2");
 final boolean formatIdentified =   rhyme.matches(formatRegex);
System.out.println(formatIdentified);//returns false

我期望的價值

"Humpty Dumpty had a great fall
 ssdsds
"

更正后的正則表達式應使用格式:abc//xxxx ，輸出應為xxxx 。

Answer 1

String.replaceAll和String.matches完全有可能以不同的方式處理多行字符串的終止符，即換行符與字符串的結尾，這就是為什么value可以打印預期結果但matches結果顯示false的原因。

我將更加明確，直接使用Pattern和Matcher而不是通過String進行代理：

        String rhyme =
                ":Humpty Dumpty sat on a wall\n" +
                "//Humpty Dumpty had a great fall\n" +
                "ssdsds\n";

        Pattern pattern = Pattern.compile("^:(.*?)//(.*[\\s\\S]?)", Pattern.DOTALL);
        Matcher matcher = pattern.matcher(rhyme);
        if (matcher.find()) {
            System.out.println(matcher.group(2));
        } else {
            System.out.println("[Pattern not found]");
        }

這將輸出：

   Humpty Dumpty had a great fall
   ssdsds

如果期望只匹配一個新行結尾，則只需將標志更改為Pattern.MULTILINE。

   Pattern pattern = Pattern.compile("^:(.*?)//(.*[\\s\\S]?)", Pattern.MULTILINE);

將輸出：

   Humpty Dumpty had a great fall

Answer 2

這將提供您想要的輸出。

      // ignore everything up to // and then include // and all following
      // in capture group 1.
      final String regex = ".*(//.*)";
      String line1 = ":Humpty Dumpty sat on a wall";
      String line2 = "//Humpty Dumpty had a great fall";
      String rhyme = line1 + line2 + "\n" + "ssdsds" + "\n";
      final String value = rhyme.replaceAll(regex, "$1");
      System.out.println(value);
          // or the follwing if you want the double quotes.
      System.out.println("\"" + value + "\"");