Question

如何使用JSON.parse reviver方法编辑某个值。我只想编辑声明为lastname的每个键，然后返回新值。

var myObj = new Object();
myObj.firstname = "mike";
myObj.lastname = "smith";

var jsonString = JSON.stringify(myObj);
var jsonObj = JSON.parse(jsonString, dataReviver);

function dataReviver(key, value)
{
    if(key == 'lastname')
    {
        var newLastname = "test";
        return newLastname;
    }
}

Answer 1

检查特殊情况后，您只需要默认传回未修改的值：

var myObj = new Object();
myObj.firstname = "mike";
myObj.lastname = "smith";

var jsonString = JSON.stringify(myObj);
var jsonObj = JSON.parse(jsonString, dataReviver);

function dataReviver(key, value)
{ 
    if(key == 'lastname')
    {
        var newLastname = "test";
        return newLastname;
    }

  return value;  // < here is where un-modified key/value pass though

}

JSON.stringify(jsonObj )// "{"firstname":"mike","lastname":"test"}"

JSON.parse使用reviver函数

1 个答案: