How do I initialize a TypeScript object with a JSO

问题:

I receive a JSON object from an AJAX call to a REST server. This object has property names that match my TypeScript class (this is a follow-on to this question).

What is the best way to initialize it? I don\'t think this will work because the class (& JSON object) have members that are lists of objects and members that are classes, and those classes have members that are lists and/or classes.

But I\'d prefer an approach that looks up the member names and assigns them across, creating lists and instantiating classes as needed, so I don\'t have to write explicit code for every member in every class (there\'s a LOT!)

回答1:

These are some quick shots at this to show a few different ways. They are by no means \"complete\" and as a disclaimer, I don\'t think it\'s a good idea to do it like this. Also the code isn\'t too clean since I just typed it together rather quickly.

Also as a note: Of course deserializable classes need to have default constructors as is the case in all other languages where I\'m aware of deserialization of any kind. Of course, Javascript won\'t complain if you call a non-default constructor with no arguments, but the class better be prepared for it then (plus, it wouldn\'t really be the \"typescripty way\").

Option #1: No run-time information at all

The problem with this approach is mostly that the name of any member must match its class. Which automatically limits you to one member of same type per class and breaks several rules of good practice. I strongly advise against this, but just list it here because it was the first \"draft\" when I wrote this answer (which is also why the names are \"Foo\" etc.).

module Environment {
    export class Sub {
        id: number;
    }

    export class Foo {
        baz: number;
        Sub: Sub;
    }
}

function deserialize(json, environment, clazz) {
    var instance = new clazz();
    for(var prop in json) {
        if(!json.hasOwnProperty(prop)) {
            continue;
        }

        if(typeof json[prop] === \'object\') {
            instance[prop] = deserialize(json[prop], environment, environment[prop]);
        } else {
            instance[prop] = json[prop];
        }
    }

    return instance;
}

var json = {
    baz: 42,
    Sub: {
        id: 1337
    }
};

var instance = deserialize(json, Environment, Environment.Foo);
console.log(instance);

Option #2: The name property

To get rid of the problem in option #1, we need to have some kind of information of what type a node in the JSON object is. The problem is that in Typescript, these things are compile-time constructs and we need them at runtime – but runtime objects simply have no awareness of their properties until they are set.

One way to do it is by making classes aware of their names. You need this property in the JSON as well, though. Actually, you only need it in the json:

module Environment {
    export class Member {
        private __name__ = \"Member\";
        id: number;
    }

    export class ExampleClass {
        private __name__ = \"ExampleClass\";

        mainId: number;
        firstMember: Member;
        secondMember: Member;
    }
}

function deserialize(json, environment) {
    var instance = new environment[json.__name__]();
    for(var prop in json) {
        if(!json.hasOwnProperty(prop)) {
            continue;
        }

        if(typeof json[prop] === \'object\') {
            instance[prop] = deserialize(json[prop], environment);
        } else {
            instance[prop] = json[prop];
        }
    }

    return instance;
}

var json = {
    __name__: \"ExampleClass\",
    mainId: 42,
    firstMember: {
        __name__: \"Member\",
        id: 1337
    },
    secondMember: {
        __name__: \"Member\",
        id: -1
    }
};

var instance = deserialize(json, Environment);
console.log(instance);

Option #3: Explicitly stating member types

As stated above, the type information of class members is not available at runtime – that is unless we make it available. We only need to do this for non-primitive members and we are good to go:

interface Deserializable {
    getTypes(): Object;
}

class Member implements Deserializable {
    id: number;

    getTypes() {
        // since the only member, id, is primitive, we don\'t need to
        // return anything here
        return {};
    }
}

class ExampleClass implements Deserializable {
    mainId: number;
    firstMember: Member;
    secondMember: Member;

    getTypes() {
        return {
            // this is the duplication so that we have
            // run-time type information :/
            firstMember: Member,
            secondMember: Member
        };
    }
}

function deserialize(json, clazz) {
    var instance = new clazz(),
        types = instance.getTypes();

    for(var prop in json) {
        if(!json.hasOwnProperty(prop)) {
            continue;
        }

        if(typeof json[prop] === \'object\') {
            instance[prop] = deserialize(json[prop], types[prop]);
        } else {
            instance[prop] = json[prop];
        }
    }

    return instance;
}

var json = {
    mainId: 42,
    firstMember: {
        id: 1337
    },
    secondMember: {
        id: -1
    }
};

var instance = deserialize(json, ExampleClass);
console.log(instance);

Option #4: The verbose, but neat way

Update 01/03/2016: As @GameAlchemist pointed out in the comments, as of Typescript 1.7, the solution described below can be written in a better way using class/property decorators.

Serialization is always a problem and in my opinion, the best way is a way that just isn\'t the shortest. Out of all the options, this is what I\'d prefer because the author of the class has full control over the state of deserialized objects. If I had to guess, I\'d say that all other options, sooner or later, will get you in trouble (unless Javascript comes up with a native way for dealing with this).

Really, the following example doesn\'t do the flexibility justice. It really does just copy the class\'s structure. The difference you have to keep in mind here, though, is that the class has full control to use any kind of JSON it wants to control the state of the entire class (you could calculate things etc.).

interface Serializable<T> {
    deserialize(input: Object): T;
}

class Member implements Serializable<Member> {
    id: number;

    deserialize(input) {
        this.id = input.id;
        return this;
    }
}

class ExampleClass implements Serializable<ExampleClass> {
    mainId: number;
    firstMember: Member;
    secondMember: Member;

    deserialize(input) {
        this.mainId = input.mainId;

        this.firstMember = new Member().deserialize(input.firstMember);
        this.secondMember = new Member().deserialize(input.secondMember);

        return this;
    }
}

var json = {
    mainId: 42,
    firstMember: {
        id: 1337
    },
    secondMember: {
        id: -1
    }
};

var instance = new ExampleClass().deserialize(json);
console.log(instance);

回答2:

TLDR: TypedJSON (working proof of concept)

---

The root of the complexity of this problem is that we need to deserialize JSON at runtime using type information that only exists at compile time. This requires that type-information is somehow made available at runtime.

Fortunately, this can be solved in a very elegant and robust way with decorators and ReflectDecorators:

1. Use property decorators on properties which are subject to serialization, to record metadata information and store that information somewhere, for example on the class prototype
2. Feed this metadata information to a recursive initializer (deserializer)

 

Recording Type-Information

With a combination of ReflectDecorators and property decorators, type information can be easily recorded about a property. A rudimentary implementation of this approach would be:

function JsonMember(target: any, propertyKey: string) {
    var metadataFieldKey = \"__propertyTypes__\";

    // Get the already recorded type-information from target, or create
    // empty object if this is the first property.
    var propertyTypes = target[metadataFieldKey] || (target[metadataFieldKey] = {});

    // Get the constructor reference of the current property.
    // This is provided by TypeScript, built-in (make sure to enable emit
    // decorator metadata).
    propertyTypes[propertyKey] = Reflect.getMetadata(\"design:type\", target, propertyKey);
}

For any given property, the above snippet will add a reference of the constructor function of the property to the hidden __propertyTypes__ property on the class prototype. For example:

class Language {
    @JsonMember // String
    name: string;

    @JsonMember// Number
    level: number;
}

class Person {
    @JsonMember // String
    name: string;

    @JsonMember// Language
    language: Language;
}

And that\'s it, we have the required type-information at runtime, which can now be processed.

 

Processing Type-Information

We first need to obtain an Object instance using JSON.parse -- after that, we can iterate over the entires in __propertyTypes__ (collected above) and instantiate the required properties accordingly. The type of the root object must be specified, so that the deserializer has a starting-point.

Again, a dead simple implementation of this approach would be:

function deserialize<T>(jsonObject: any, Constructor: { new (): T }): T {
    if (!Constructor || !Constructor.prototype.__propertyTypes__ || !jsonObject || typeof jsonObject !== \"object\") {
        // No root-type with usable type-information is available.
        return jsonObject;
    }

    // Create an instance of root-type.
    var instance: any = new Constructor();

    // For each property marked with @JsonMember, do...
    Object.keys(Constructor.prototype.__propertyTypes__).forEach(propertyKey => {
        var PropertyType = Constructor.prototype.__propertyTypes__[propertyKey];

        // Deserialize recursively, treat property type as root-type.
        instance[propertyKey] = deserialize(jsonObject[propertyKey], PropertyType);
    });

    return instance;
}

var json = \'{ \"name\": \"John Doe\", \"language\": { \"name\": \"en\", \"level\": 5 } }\';
var person: Person = deserialize(JSON.parse(json), Person);

The above idea has a big advantage of deserializing by expected types (for complex/object values), instead of what is present in the JSON. If a Person is expected, then it is a Person instance that is created. With some additional security measures in place for primitive types and arrays, this approach can be made secure, that resists any malicious JSON.

 

Edge Cases

However, if you are now happy that the solution is that simple, I have some bad news: there is a vast number of edge cases that need to be taken care of. Only some of which are:

• Arrays and array elements (especially in nested arrays)
• Polymorphism
• Abstract classes and interfaces
• ...

If you don\'t want to fiddle around with all of these (I bet you don\'t), I\'d be glad to recommend a working experimental version of a proof-of-concept utilizing this approach, TypedJSON -- which I created to tackle this exact problem, a problem I face myself daily.

Due to how decorators are still being considered experimental, I wouldn\'t recommend using it for production use, but so far it served me well.

回答3:

you can use Object.assign I don\'t know when this was added, I\'m currently using Typescript 2.0.2, and this appears to be an ES6 feature.

client.fetch( \'\' ).then( response => {
        return response.json();
    } ).then( json => {
        let hal : HalJson = Object.assign( new HalJson(), json );
        log.debug( \"json\", hal );

here\'s HalJson

export class HalJson {
    _links: HalLinks;
}

export class HalLinks implements Links {
}

export interface Links {
    readonly [text: string]: Link;
}

export interface Link {
    readonly href: URL;
}

here\'s what chrome says it is

HalJson {_links: Object}
_links
:
Object
public
:
Object
href
:
\"http://localhost:9000/v0/public

so you can see it doesn\'t do the assign recursively

回答4:

I\'ve been using this guy to do the job: https://github.com/weichx/cerialize

It\'s very simple yet powerful. It supports:

• Serialization & deserialization of a whole tree of objects.
• Persistent & transient properties on the same object.
• Hooks to customize the (de)serialization logic.
• It can (de)serialize into an existing instance (great for Angular) or generate new instances.
• etc.

Example:

class Tree {
  @deserialize public species : string; 
  @deserializeAs(Leaf) public leafs : Array<Leaf>;  //arrays do not need extra specifications, just a type.
  @deserializeAs(Bark, \'barkType\') public bark : Bark;  //using custom type and custom key name
  @deserializeIndexable(Leaf) public leafMap : {[idx : string] : Leaf}; //use an object as a map
}

class Leaf {
  @deserialize public color : string;
  @deserialize public blooming : boolean;
  @deserializeAs(Date) public bloomedAt : Date;
}

class Bark {
  @deserialize roughness : number;
}

var json = {
  species: \'Oak\',
  barkType: { roughness: 1 },
  leafs: [ {color: \'red\', blooming: false, bloomedAt: \'Mon Dec 07 2015 11:48:20 GMT-0500 (EST)\' } ],
  leafMap: { type1: { some leaf data }, type2: { some leaf data } }
}
var tree: Tree = Deserialize(json, Tree);

回答5:

I\'ve created a tool that generates TypeScript interfaces and a runtime \"type map\" for performing runtime typechecking against the results of JSON.parse: ts.quicktype.io

For example, given this JSON:

{
  \"name\": \"David\",
  \"pets\": [
    {
      \"name\": \"Smoochie\",
      \"species\": \"rhino\"
    }
  ]
}

quicktype produces the following TypeScript interface and type map:

export interface Person {
    name: string;
    pets: Pet[];
}

export interface Pet {
    name:    string;
    species: string;
}

const typeMap: any = {
    Person: {
        name: \"string\",
        pets: array(object(\"Pet\")),
    },
    Pet: {
        name: \"string\",
        species: \"string\",
    },
};

Then we check the result of JSON.parse against the type map:

export function fromJson(json: string): Person {
    return cast(JSON.parse(json), object(\"Person\"));
}

I\'ve left out some code, but you can try quicktype for the details.

回答6:

Option #5: Using Typescript constructors and jQuery.extend

This seems to be the most maintainable method: add a constructor that takes as parameter the json structure, and extend the json object. That way you can parse a json structure into the whole application model.

There is no need to create interfaces, or listing properties in constructor.

export class Company
{
    Employees : Employee[];

    constructor( jsonData: any )
    {
        jQuery.extend( this, jsonData);

        // apply the same principle to linked objects:
        if ( jsonData.Employees )
            this.Employees = jQuery.map( jsonData.Employees , (emp) => {
                return new Employee ( emp );  });
    }

    calculateSalaries() : void { .... }
}

export class Employee
{
    name: string;
    salary: number;
    city: string;

    constructor( jsonData: any )
    {
        jQuery.extend( this, jsonData);

        // case where your object\'s property does not match the json\'s:
        this.city = jsonData.town;
    }
}

In your ajax callback where you receive a company to calculate salaries:

onReceiveCompany( jsonCompany : any ) 
{
   let newCompany = new Company( jsonCompany );

   // call the methods on your newCompany object ...
   newCompany.calculateSalaries()
}

回答7:

The 4th option described above is a simple and nice way to do it, which has to be combined with the 2nd option in the case where you have to handle a class hierarchy like for instance a member list which is any of a occurences of subclasses of a Member super class, eg Director extends Member or Student extends Member. In that case you have to give the subclass type in the json format

回答8:

Maybe not actual, but simple solution:

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

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

work for difficult dependencies too!!!

回答9:

JQuery .extend does this for you:

var mytsobject = new mytsobject();

var newObj = {a:1,b:2};

$.extend(mytsobject, newObj); //mytsobject will now contain a & b

回答10:

Another option using factories

export class A {

    id: number;

    date: Date;

    bId: number;
    readonly b: B;
}

export class B {

    id: number;
}

export class AFactory {

    constructor(
        private readonly createB: BFactory
    ) { }

    create(data: any): A {

        const createB = this.createB.create;

        return Object.assign(new A(),
            data,
            {
                get b(): B {

                    return createB({ id: data.bId });
                },
                date: new Date(data.date)
            });
    }
}

export class BFactory {

    create(data: any): B {

        return Object.assign(new B(), data);
    }
}

https://github.com/MrAntix/ts-deserialize

use like this

import { A, B, AFactory, BFactory } from \"./deserialize\";

// create a factory, simplified by DI
const aFactory = new AFactory(new BFactory());

// get an anon js object like you\'d get from the http call
const data = { bId: 1, date: \'2017-1-1\' };

// create a real model from the anon js object
const a = aFactory.create(data);

// confirm instances e.g. dates are Dates 
console.log(\'a.date is instanceof Date\', a.date instanceof Date);
console.log(\'a.b is instanceof B\', a.b instanceof B);

1. keeps your classes simple
2. injection available to the factories for flexibility

回答11:

For simple objects, I like this method:

class Person {
  constructor(
    public id: String, 
    public name: String, 
    public title: String) {};

  static deserialize(input:any): Person {
    return new Person(input.id, input.name, input.title);
  }
}

var person = Person.deserialize({id: \'P123\', name: \'Bob\', title: \'Mr\'});

Leveraging the ability to define properties in the constructor lets it be concise.

This gets you a typed object (vs all the answers that use Object.assign or some variant, which give you an Object) and doesn\'t require external libraries or decorators.

回答12:

you can do like below

export interface Instance {
  id?:string;
  name?:string;
  type:string;
}

and

var instance: Instance = <Instance>({
      id: null,
      name: \'\',
      type: \'\'
    });

回答13:

**model.ts**
export class Item {
    private key: JSON;
    constructor(jsonItem: any) {
        this.key = jsonItem;
    }
}

**service.ts**
import { Item } from \'../model/items\';

export class ItemService {
    items: Item;
    constructor() {
        this.items = new Item({
            \'logo\': \'Logo\',
            \'home\': \'Home\',
            \'about\': \'About\',
            \'contact\': \'Contact\',
        });
    }
    getItems(): Item {
        return this.items;
    }
}