Typescript objects serialization?

Question

Are there any means for JSON serialization/deserialization of Typescript objects so that they don't loose type information? Simple JSON.parse(JSON.stringify) has too many caveats.

Or I should use adhoc solutions?

Answer 1

Use Interfaces to get strong types:

// 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); 

And use type assertion "<>" if you need to.

Answer 2

I think a better way to handle this is to use Object.assign (which however requires ECMAScript 2015).

Given a class

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);     } } 

Serialize and deserialize like this...

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

To take this example to the next level, consider nested objects...

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;     } } 

Serialize and deserialize like this...

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()); 

So unlike using an interface, this approach preserves methods.