Question

当JSON数据具有嵌套对象时，我无法将JSON转换为Javascript对象。顶级＆＃39;人物＆＃39;对象得到了很好的重建，但是＆＃39; Residence＆＃39;对象属性不是

function Person(first, last) {
    this.FirstName = first;
    this.LastName = last;
    this.Residence = {};
}

Person.Revive = function (data) {
    return new Person(data.FirstName, data.LastName);
}

Object.defineProperty(Person.prototype, "FullName", {
    get: function() { return this.FirstName + " " + this.LastName; }
});

Person.prototype.toJSON = function () {
   this.__class__ = "Person";
    return this;
});

function Residence(lat, long) {
    this.Latitude = lat;
    this.Longitude = long;
}

Residence.prototype.toJSON = function () {
    this.__class__ = "Residence";
    return this;
}

Residence.Revive = function (data) {
    return new Residence(data.Latitude, data.Longitude);
}

Object.defineProperty(Residence.prototype, "Location", {
    get: function () { return this.Latitude + ", " + this.Longitude; }
});

var p = new Person("Foo", "Bar");
p.Residence = new Residence(44, 33);

console.log("Full name = " + p.FullName);
console.log("Location = " + p.Residence.Location);

var serialization = JSON.stringify(p);
console.log(serialization);

var rawObj = JSON.parse(serialization, function (key, value) {
    if (value instanceof Object && value.__class__ == 'Person') {
        return Person.Revive(value);
    }
    if (value instanceof Object && value.__class__ == 'Residence') {
        return Residence.Revive(value);
    }
    return value;
});
console.log("Full name = " + rawObj.FullName);
console.log("Location = " + rawObj.Residence.Location);

JSON.parse函数确实为＆＃39; Residence＆＃39;提供了一个键/值对。对象，并创建并返回一个新的Residence对象。然而，由此产生的＆lt; rawObj.Residence＆＃39;只是一个空对象。谁能指出我做错了什么？

控制台输出如下：

Full name = Foo Bar
Location = 44, 33
{"FirstName":"Foo","LastName":"Bar","Age":22,"Residence":{"Latitude":44,"Longitude":33,"__class__":"Residence"},"__class__":"Person"}
Full name = Foo Bar
Location = undefined

小提琴：http://jsfiddle.net/CadGuy/yyq4dqtx/

Answer 1

var p = new Person("Foo", "Bar");
p.Residence = new Residence(44, 33);

好吧，如果你正在构建这样的Person个对象，你也必须像这样重振它们：

Person.Revive = function (data) {
    var p = new Person(data.FirstName, data.LastName);
    p.Residence = data.Residence;
    return p;
};

当然，首先将住所作为Person的{可选？）参数可能是个好主意。

JSON.parse（），重新连接到嵌套对象

1 个答案: