Javascript regex替换字符串中的所有双引号，除非双引号后面跟着空格或逗号空格

这是我能做到的最好的，利用JSON中未转义引号只能出现在某些地方的事实。

input = '{"description":"This is my 12" pizza, and I want "thin crust"","value":"1"}';
console.log(input);
output = input.replace(/{"/g, '_OPEN_').replace(/":"/g, '_COLON_').replace(/","/g, '_COMMA_').replace(/"}/g, '_CLOSE_');
output = output.replace(/"/g, '''"');
output = output.replace(/_OPEN_/g, '{"').replace(/_COLON_/g, '":"').replace(/_COMMA_/g, '","').replace(/_CLOSE_/g, '"}');
console.log(output)

生产

{"description":"This is my 12" pizza, and I want "thin crust"","value":"1"}
{"description":"This is my 12'" pizza, and I want '"thin crust'"","value":"1"}

您可以将'OPEN'， 'CLOSE'等替换为不太可能在输入中出现的字符串，如果您不介意正则表达式是隐式的，甚至可能是控制字符。但正如其他人所指出的，没有一种解决方案在所有情况下都有效。无论您做什么，描述文本中都有可能出现一个值，它会把您搞得一团糟，因为与正确生成的JSON不同，您试图解析的语法是不明确的。

试试这个替换:

repl = str.replace(/"(?= )/g, "'''"");

不是一个单行正则表达式，但我认为这样做更安全:

 json_string = '{"description":"This is my 12" pizza: which can also contain other "," which would break in a one liner regex.","value":"1"}';
 console.log(json_string);
 // save the value for later use
 var value = json_string.match(/"value":"(.+)"}$/)[1];
 // isolate just the description value..
 // remove the ","value... from the end
 var desc = json_string.replace(/","value":".+"}$/, '');
 // remove the opening {"description":" from the description value
 desc = desc.replace(/^{"description":"/, '');
 // any remaining " in the description are unwanted to replace them
 desc = desc.replace(/"/g, '"');
 console.log(desc);
 // now put it all back together - if you wanted too - but really you already have the description and value parsed out of the string
 json_string = '{"description":"'+desc+'","value":"'+value+'"}'
 console.log(json_string);

控制台输出如下:

{"description":"This is my 12" pizza: which can also contain other "," which would break in a one liner regex.","value":"1"}
This is my 12" pizza: which can also contain other "," which would break in a one liner regex.
{"description":"This is my 12" pizza: which can also contain other "," which would break in a one liner regex.","value":"1"}

注意这个方法不会破坏，如果描述也包含任何模式，你可能会使用在一个正则表达式的一部分

一种重建json字符串的方法:

var str = '{"description":"This is my 12" pizza I ordered.","value":"1"}';
var regex = /"(.*?)"(?='s*([,:])'s*"|(}))/g;
var result = '{';
var arr = regex.exec(str);
while (arr != null) {
    result += '"' + arr[1].replace(/''?"/g, '''"') + '"';
    if (arr[2]) result += arr[2];
    if (arr[3]) result += arr[3];
    arr = regex.exec(str);
}
console.log(result);

我讨厌回答我自己的问题，正如你们许多人指出的那样，正确地，在事实之后修复错误的JSON是不可能的，因为一个不可靠的、未转义的双引号字符注入了歧义。也许这应该是一个正确的答案，不幸的是它不能解决我的问题。

对于那些遇到同样问题的人-我希望下面的函数可以帮助解决问题，直到您可以获得错误JSON的来源。本质上，您必须查看每个双引号，然后查看两个字符(至少)的前后，并根据前/下一个字符评估它在JSON中是否可能有效使用双引号或无效。如果无效，则插入转义字符。下面的函数在这方面做得很好，不过，根据双引号导致的JSON的畸形程度，您可能需要稍微扩展一下这个函数。我的希望是，对于那些在我的情况下，这至少提供了一个良好的起点。

感谢所有的贡献-潜在解决方案的数量和广度是相当了不起的回顾!

// fix unescaped double quotes / malformed JSON
function cleanseJSON(jsonStr)
{
  for(var k=0;k<jsonStr.length;k++)
  { 
    if(jsonStr.charAt(k)=='"')
    {
      var prevChar=jsonStr.charAt(k-1);
      var prevChar2=jsonStr.charAt(k-2);
      var nextChar=jsonStr.charAt(k+1);
      var nextChar2=jsonStr.charAt(k+2);
      var esc="''";
      var isValid=false;
      var prevFix=false;
      var postFix=false;
      switch(prevChar)
      {
        case ':':
        case '{':
        case ',':
        case '[':
        case '''':   // already escaped
          isValid=true;
          break;
        default:
          prevFix=true;
      }
      switch(nextChar)
      {
        case ':':
        case '}':
        case ',':
          if(nextChar2==' '){   // if there is a comma, but the next is a space consider it invalid JSON
            break;  
          }
        case ']':
        case '''':   // already escaped
          isValid=true;
          break;
        default:
          postFix=true;
      }
      // first test to ensure the quote is likely bogus
      if(!isValid)
      {
        if(prevFix){
          jsonStr = [jsonStr.slice(0, k), esc, jsonStr.slice(k)].join('');   
        } else {
          if(postFix){
            jsonStr = [jsonStr.slice(0, k+1), esc, jsonStr.slice(k+1)].join('');
          }
        }
      } // if not valid "
    } // if we find a doublequote
  } // for each char in the jsonStr
  return jsonStr;
}