我怎样才能捕获所有组并将它们替换为组的重新格式化版本？

Question

文本输入：

some "single , quote" , text ""{one,2two space,three-dash,four}"" "{some,text}"  ""{alpha,bravo space - dash,charlie}"" some text

文本输出：

some "single , quote" , text ""{one","2two space","three-dash","four}"" "{some,text}"  ""{alpha","bravo space - dash","charlie}"" some text

我有下面的javascript解决方案，但我想知道是否有更好的解决方案？

const str = "some \"single , quote\" , text \"\"\{one,2two space,three-dash,four\}\"\" \"\{some,text\}\"  \"\"\{alpha,bravo space - dash,charlie\}\"\" some text"
let res = str;
const matches = res.match(/""{([^"])*}""/g);
matches?.forEach( match => {
   res = res.replace(match, match.replace(/,/g,'","'));
});
console.log(str); // TEXT IN
console.log(res); // TEXT OUT

Answer 1

你可以给一个函数作为 replace() 方法的替换参数，这样你就不需要循环了。

此外，您的正则表达式可以简化。您不需要捕获组，只需将 * 量词放在 [^"]

之后

const str = 'some, text ""{one,2two space,three-dash,four}"" some, text ""{alpha,bravo space - dash,charlie}"" some text';

res = str.replace(/""{[^"]*}""/g, match => match.replace(/,/g, '","'))
console.log(res);

Answer 2

您可以使用以下正则表达式来匹配要替换为（三字符字符串）'","'.

的逗号

/(?<=""{[^{}\n]*),(?=[^{}\n]*}"")/

Start your engine!

这是可能的，因为 Javascript 支持 可变长度 回顾。

对于字符串

some, text ""{one,2two space,three-dash,four}"" some, text ""{alpha,bravo space - dash,charlie}"" some text';

字符串中的匹配和替换结果

some, text ""{one","2two space","three-dash","four}"" some, text ""{alpha","bravo space - dash","charlie}"" some text';

Javascript 的正则表达式引擎执行以下操作。

(?<=        : begin a positive lookbehind
  ""{       : match '""{'
  [^{}\n]*  : match 0+ chars other than those shown in the char class
)           : end positive lookbehind
,           : match ','
(?=         : begin positive lookahead
  [^{}\n]*  : match 0+ chars other than those shown in the char class
  }""       : match '}""'
)           : end positive lookahead

我怎样才能捕获所有组并将它们替换为组的重新格式化版本？

How can I capture all groups and replace them with the reformatted version of the group?

javascript

regex

regex-group