打字稿对象序列化？

Question

是否有任何方法可以对Typescript对象进行JSON序列化/反序列化,以便它们不会丢失类型信息？简单JSON.parse(JSON.stringify)有太多警告.

或者我应该使用adhoc解决方案？

Answer 1

使用接口获取强类型:

// Creating 
var foo:any = {};
foo.x = 3;
foo.y='123';

var jsonString = JSON.stringify(foo);
alert(jsonString);


// Reading
interface Bar{
    x:number;
    y?:string; 
}

var baz:Bar = JSON.parse(jsonString);
alert(baz.y);

如果需要,可以使用类型断言"<>".

Answer 2

我认为处理此问题的更好方法是使用Object.assign（但是需要ECMAScript 2015）。

上一堂课

class Pet {
    name: string;
    age: number;
    constructor(name?: string, age?: number) {
        this.name = name;
        this.age = age;
    }
    getDescription(): string {
        return "My pet " + this.name + " is " + this.age + " years old.";
    }
    static fromJSON(d: Object): Pet {
        return Object.assign(new Pet(), d);
    }
}

像这样序列化和反序列化...

var p0 = new Pet("Fido", 5);
var s = JSON.stringify(p0);
var p1 = Pet.fromJSON(JSON.parse(s));
console.log(p1.getDescription());

为了使这个例子更上一层楼，请考虑嵌套对象...

class Type {
    kind: string;
    breed: string;
    constructor(kind?: string, breed?: string) {
        this.kind = kind;
        this.breed = breed;
    }
    static fromJSON(d: Object) {
        return Object.assign(new Type(), d);
    }
}
class Pet {
    name: string;
    age: number;
    type: Type;
    constructor(name?: string, age?: number) {
        this.name = name;
        this.age = age;
    }
    getDescription(): string {
        return "My pet " + this.name + " is " + this.age + " years old.";
    }
    getFullDescription(): string {
        return "My " + this.type.kind + ", a " + this.type.breed + ", is " + this.age + " years old.";
    }
    static fromJSON(d: Object): Pet {
        var o = Object.assign(new Pet(), d);
        o.type = Type.fromJSON(o['type']);
        return o;
    }
}

像这样序列化和反序列化...

var q0 = new Pet("Fido", 5);
q0.type = new Type("dog", "Pomeranian");
var t = JSON.stringify(q0);
var q1 = Pet.fromJSON(JSON.parse(t));
console.log(q1.getFullDescription());

因此，与使用接口不同，此方法保留了方法。