替换{和}

Question

有一个问题。 在JSON.stringify之后，我有一个字符串：

value.value_new = {"apiunits":{"amount":"0"},"total":{"requests":"10","results":"10"},"project":{"projects":"1"}};

我想替换{和}，然后尝试通过以下方式进行操作：

value.value_new = value.value_new.replace("/[{}]/g", " ");

要么

value.value_new = value.value_new.replace("/{/g", " ");
value.value_new = value.value_new.replace("/}/g", " ");

但这是行不通的。 为什么？

Answer 1

我不确定为什么要这样做或对结果会做什么，但是我认为您想要的RegEx是：

var asString = '{"apiunits":{"amount":"0"},"total":{"requests":"10","results":"10"},"project":{"projects":"1"}};'

//Replace either curly with a space. Note in RegEx you have to escape the curly braces with a backslash
var replaced = asString.replace(/\{|\}/gi, ' ');

console.log(replaced);
//outputs -->" "apiunits": "amount":"0" ,"total": "requests":"10","results":"10" ,"project": "projects":"1"  ;"

在我的示例中，我将管道用于“或”，但是如果花括号被转义，则您先前的任何尝试都将起作用：

value = value.replace("/[\{\}]/g", " ");

要么

 value = value.replace("/\{/g", " ");
 value = value.replace("/\}/g", " ");

Answer 2

value.value_new不是字符串，因此您不能在其上使用函数replace 。

如果只需要不带括号的此字符串，则可以使用以下方法：

var value = {};
value.value_new = '{"apiunits":{"amount":"0"},"total":{"requests":"10","results":"10"},"project":{"projects":"1"}}';

String.prototype.replaceAll = function(str1, str2, ignore) 
{
    return this.replace(new RegExp(str1.replace(/([\/\,\!\\\^\$\{\}\[\]\(\)\.\*\+\?\|\<\>\-\&])/g,"\\$&"),(ignore?"gi":"g")),(typeof(str2)=="string")?str2.replace(/\$/g,"$$$$"):str2);
} 

value.value_new = value.value_new.replaceAll("{", '');
value.value_new = value.value_new.replaceAll("}", '');


console.log(value.value_new); // You'll have the string that you want