JAVA中带反向引用的递归组捕获正则表达式

Question

我试图在字符串中递归地捕获多个组，也使用反向引用到正则表达式内的组。即使我使用Pattern和Matcher以及"while(Matcher .find())"循环，它仍然只捕获最后一个实例，而不是所有实例。在我的例子中，唯一可能的标记是，，，，，，，，。因为这些是格式化标记，所以我需要捕获:

1. 标签之外的任何文本(以便我可以将其格式化为"正常"文本，并且我通过在一组中捕获标签本身的同时在另一组中捕获标签本身来捕获任何文本，并且当我迭代出现时，我删除了从原始字符串中捕获的所有内容;如果最后还有剩余的文本，我将其格式化为"normal"文本)
2. 标签的"名称"，以便我知道我将如何拥有设置
标签内的文本格式
3. 标签的文本内容，将根据标签名称及其相关规则进行格式化

下面是我的示例代码:

        String currentText = "the man said:<pof>“This one, at last, is bone of my bones</pof><poi>and flesh of my flesh;</poi><po>This one shall be called ‘woman,’</po><poil>for out of man this one has been taken.”</poil>";
        String remainingText = currentText;
        //first check if our string even has any kind of xml tag, because if not we will just format the whole string as "normal" text
        if(currentText.matches("(?su).*<[/]{0,1}(?:sm|po)[f|l|s|i|3]{0,1}[f|l]{0,1}>.*"))
        {                
            //an opening or closing tag has been found, so let us start our pattern captures
            //I am using a backreference \2 to make sure the closing tag is the same as the opening tag
            Pattern pattern1 = Pattern.compile("(.*)<((sm|po)[f|l|s|i|3]{0,1}[f|l]{0,1})>(.*?)</\2>",Pattern.UNICODE_CHARACTER_CLASS);
            Matcher matcher1 = pattern1.matcher(currentText);                
            int iteration = 0;
            while(matcher1.find()){
                System.out.print("Iteration ");
                System.out.println(++iteration);
                System.out.println("group1:"+matcher1.group(1));
                System.out.println("group2:"+matcher1.group(2));
                System.out.println("group3:"+matcher1.group(3));
                System.out.println("group4:"+matcher1.group(4));
                if(matcher1.group(1) != null && matcher1.group(1).isEmpty() == false)
                {
                    m_xText.insertString(xTextRange, matcher1.group(1), false);
                    remainingText = remainingText.replaceFirst(matcher1.group(1), "");
                }
                if(matcher1.group(4) != null && matcher1.group(4).isEmpty() == false)
                {
                    switch (matcher1.group(2)) {
                        case "pof": [...]
                        case "pos": [...]
                        case "poif": [...]
                        case "po": [...]
                        case "poi": [...]
                        case "pol": [...]
                        case "poil": [...]
                        case "sm": [...]
                    }
                    remainingText = remainingText.replaceFirst("<"+matcher1.group(2)+">"+matcher1.group(4)+"</"+matcher1.group(2)+">", "");
                }
            }

System.out.println只在控制台中输出一次，结果如下:

Iteration 1:
  group1:the man said:<pof>“This one, at last, is bone of my bones</pof><poi>and flesh of my flesh;</poi><po>This one shall be called ‘woman,’</po>; 
  group2:poil
  group3:po
  group4:for out of man this one has been taken.”

第3组将被忽略，唯一有用的组是1、2和4(第3组是第2组的一部分)。为什么这只捕获最后一个标签实例"poil"，而不捕获前面的"poof"、"poi"one_answers"po"标签?

我希望看到的输出是这样的:

Iteration 1:
  group1:the man said:
  group2:pof
  group3:po
  group4:“This one, at last, is bone of my bones
Iteration 2:
  group1:
  group2:poi
  group3:po
  group4:and flesh of my flesh;
Iteration 3:
  group1:
  group2:po
  group3:po
  group4:This one shall be called ‘woman,’
Iteration 3:
  group1:
  group2:poil
  group3:po
  group4:for out of man this one has been taken.”

Answer 1

我刚刚找到了这个问题的答案，它只需要在第一个捕获中使用一个非贪婪量词，就像我在第四个捕获组中一样。这是完全按照需要工作的:

Pattern pattern1 = Pattern.compile("(.*?)<((sm|po)[f|l|s|i|3]{0,1}[f|l]{0,1})>(.*?)</\2>",Pattern.UNICODE_CHARACTER_CLASS);