1. JSON.parse(text[, reviver])
text 必需 有效的json字符串
reviver 可选 函数
2. 举栗子
1) 只有第一个参数
let objStr = ‘{"name": "lqw","age": 23,"sex": "female"}‘;
console.log(JSON.parse(objStr));
结果：
{name: "lqw", age: 23, sex: "female"}
2）第二个参数存在
let objStr = ‘{"name": "lqw","age": 23,"sex": "female"}‘;
console.log(JSON.parse(objStr, function (key, value) {
if (typeof value === ‘string‘) {
return value.toUpperCase();
}
return value;
}));
结果：
{name: "LQW", age: 23, sex: "FEMALE"}
原文：https://www.cnblogs.com/lqw007/p/9618420.html