java – 用于替换嵌套标记的正则表达式

2023年2月16日 168次阅读

我需要采用正则表达式模式并以编程方式逃避花括号.输入正则表达式将匹配以下模式(标签之前,之后和之间的文本)：

&{token1}
&{token1}&{token2}&{tokenN...}
&{token1&{token2&{tokenN...}}}

到目前为止,我对除嵌套标签之外的所有内容都很好.这就是我所拥有的.

regex = regex.replaceAll("(&)(\\{)([^{}]+)(\\})", "$1\\\\$2$3\\\\$4");

我也试过使用迭代和递归,但我遇到的问题是,一旦最内层的令牌被转义,它就会与匹配混乱.

我尝试过负面的看守,但这并不符合我的预期.它只会匹配/替换最里面的令牌.

regex = regex.replaceAll("(&)(\\{)([^(?<!\\\\{)|(?<!\\\\})]+)(\\})", "$1\\\\$2$3\\\\$4");

有什么建议？提前致谢.

编辑：输入/输出示例

&{token1}   //input
&\{token1\} //output

&{token1}&{token2}&{tokenN...}        //input
&\{token1\}&\{token2\}&\{tokenN...\}  //output

&{token1&{token2&{tokenN...}}}        //input
&{token1&{token2&\{tokenN...\}}}      //output
&\{token1&\{token2&\{tokenN...\}\}\}  //expected output

//To throw a wrench into it, normal quantifiers should not be escaped
text{1,2}&{token1&{token2&{tokenN...}}}        //input
text{1,2}&{token1&{token2&\{tokenN...\}}}      //output
text{1,2}&\{token1&\{token2&\{tokenN...\}\}\}  //expected output

编辑2：在此过程之外发生的事例的示例：标签将被解析为文本,然后最终,它应该是有效的正则表达式.

a{2}&{token1&{token2&{tokenN...}}}        //input
a{2}&\{token1&\{token2&\{tokenN...\}\}\}  //expected output of this regex
a{2}foobarbaz                             //expected output after tokens are resolved (&{token1} = foo, &{token2} = bar, &{tokenN...} = baz)

最佳答案 试试：

regex = regex.replaceAll("(?<=&)(?=\\{)|(?<!\\{\\d{0,6},?(\\d{0,6})?)(?=\\})","\\\\");

其中(0,6)确定可以有多少位数,6我认为足够了
Java示例：

public class Main {
    public static void main(String[] args){
        int i = 0;
        String regex = "&{token1}&{token2}&{tokenN}\n" +
                "&{token1&{token2&{tokenN}}}\n" +
                "text{1,2}&{token1{1}&{token2{1,}&{tokenN{0,2}}}}\n";
        regex = regex.replaceAll("(?<=&)(?=\\{)|(?<!\\{\\d{0,6},?(\\d{0,6})?)(?=\\})","\\\\");
        System.out.println(regex);
    }
}

输出：

&\{token1\}&\{token2\}&\{tokenN\}
&\{token1&\{token2&\{tokenN\}\}\}
text{1,2}&\{token1{1}&\{token2{1,}&\{tokenN{0,2}\}\}\}
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