I am using Json2Typescript to serialize and serialize typescript object. The problem is that I am not able to deserialize a JSON Data that I receive from my back-end.
This is my typescript class:
#JsonObject('Respondent')
export class Respondent {
#JsonProperty('id',Number)
Id:number;
video:string;
#JsonProperty('synchro_id',Number)
SynchroId:string;
#JsonProperty('language',String)
Language:string;
#JsonProperty('project_id',Number)
ProjectId:number;
#JsonProperty('terminal_id',Number)
TerminalId:number;
#JsonProperty('questionnaire_id',Number)
QuestionnaireId: number;
#JsonProperty('device_type_id',Number)
DeviceTypeId:string;
// TODO: This is to be removed
medias: Media[] = null;
// TODO: This is to be removed
clicks: Click[] = null;
#JsonProperty('email_status',String)
emailStatus:string;
#JsonProperty('inserted_local_time',String)
insertedTime:string;
#JsonProperty('responses',[Answer])
Answers:Answer[];// TSMap<number,Answer>; // {[key: number]: Answer[]} = {};
#JsonProperty('size_on_disk',Number)
sizeOnDisk:number;
#JsonProperty('email_opened',Boolean)
emailOpened:boolean;
#JsonProperty('token',String)
token:string;
constructor() {
}
...
}
This is how I deserialize my JSON.
private getRespondents(start:number,end:number) {
this._projectService.getRespondentsByRange(start,end,-1,-1,-1,this.questionnaire.id,-1,false).subscribe((respondents) => {
this.respondents = new Array<Respondent>();
this.respondents = new JsonConvert().deserializeArray(respondents,Respondent); // doe s not work
if(this.respondents!=null && this.respondents.length>0) {
this._sizeOnDisk = this.respondents[0].sizeOnDisk * this._nbOfRespondents;
}
},
error => this._errorMessage = <any>error);
}
Ok, I have found the answer.
It is not explicitly specified in the official doc.
YOU HAVE TO INITIALIZE THE CLASS ATTRIBUTES.
#JsonProperty("email", String)
public email: String = "";
Check this link for the answer
I know that function overloading is not supported in typescript and javascript.
And I'm studying a detour method to work like this function overloading.
My case is as follows.
first:
The last ans first argument is fixed.
public A(arg1:number, arg2:number, argLast:any){
}
public A(arg1:number, arg2:number, arg3:Array<any>, argLast:any){
}
Second :
There is an indicator for whether the function is overloaded.
Of course, as in the above example, it is possible, but I have to create it through a new interface, so it does not fit in my case.
I have tried various function overloading methods.
I also implemented it through john resig blog.
(https://johnresig.com/blog/javascript-method-overloading/)
The code below is an example I made through the above blog.
function addMethod (object, name, fn) {
var old = object[ name ]
object[ name ] = function () {
if (fn.length === arguments.length) {
return fn.apply(this, arguments)
} else if (typeof old === 'function') {
return old.apply(this, arguments)
}
}
}
export class Users{
find(...arg: any[]){
Users.prototype['findOVF'](arg)
}
}
addMethod(Users.prototype, 'findOVF', function () {
console.log('ARG 0')
})
addMethod(Users.prototype, 'findOVF', function (name:string, age:number) {
console.log('ARG 2')
})
addMethod(Users.prototype, 'findOVF', function (first_name:string, last_name:string,age:number) {
console.log('ARG 3')
})
var users = new Users()
users.find()
users.find('John',19)
users.find('John', 'Resig', 19)
When I use this method, I call the function with the parameter (... arg).
I've tried to implement all the methods of the blog below.
(https://blog.mariusschulz.com/2016/08/18/function-overloads-in-typescript)
public A(arg1:number, arg2:number, argLast:any){
}
public A(arg1:number, arg2:number, arg3:Array<any>|null, argLast:any){
}
If I implement function overloading this way, I get an error every time I call the A function. It's not really function overloading.
But I have not found a way to implement these two elements at the same time.
I want to know if this is impossible with the typescript grammar or if there is any other possibility.
Thanks for seeing my awkward English.
You can create overloads in typescript but all the overloads have a single implementation and it's up to you to differentiate between them. The simplest solution, if your methods are all differentiated just by number of parameters is this:
class User { }
export class Users {
find(): User;
find(name: string, age: number): User;
find(first_name: string, last_name: string, age: number): User
find(...args: [any?, any?, any?]): User {
if (args.length == 0) {
return null as any;
} else if (args.length == 1) {
let [name, age, _]: [string?, number?, any?] = args;
console.log(`${name}, ${age}`);
return null as any;
} else if (args.length == 2){
let [first_name, last_name, age]: [string?, string?, number?] = args;
console.log(`${first_name}, ${last_name}, ${age}`);
return null as any;
} else {
throw "Not implemented";
}
}
}
Playground link
The above solution preserves call site type safety, your solution does not since the argument to find is any[] in your example.
You could use a more automated approach by defining a function that will create an overloaded function form an array of functions but wather this extra complexity is worth it you be the judge.
type UnionToIntersection<U> =
(U extends any ? (k: U) => void : never) extends ((k: infer I) => void) ? I : never
function overloads<TThis>(){
return function<T extends Array<(this: TThis, ...args: any[]) => any>>(...fns: T): UnionToIntersection<T[number]> {
return function (this: TThis, ...args: any[]) {
for (var fn of fns) {
if (fn.length === args.length) {
return fn.apply(this, args);
}
}
throw "Not implemented";
} as any
}
}
class User { }
export class Users {
member: string = "M"
find = overloads<Users>()(
function () {
console.log('ARG 0');
this.member // this is Users
},
function (name: string, age: number) {
console.log('ARG 2')
},
function (first_name: string, last_name: string, age: number) {
console.log('ARG 3')
}
);
}
var users = new Users()
users.find('John', 19)
users.find('John', 'Resig', 19)
Playground link
Or a version that assigns the function to the prototype not the instance:
export class Users {
member: string = "M"
}
let find = overloads<Users>()(
function () {
console.log('ARG 0');
this.member // this is Users
},
function (name: string, age: number) {
console.log('ARG 2')
},
function (first_name: string, last_name: string, age: number) {
console.log('ARG 3')
}
);
export interface Users {
find: typeof find;
}
Users.prototype.find = find;
Playground link
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!)
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 (idea, implementation), 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);
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
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:
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
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.
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.
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);
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.
This is my approach (very simple):
const jsonObj: { [key: string]: any } = JSON.parse(jsonStr);
for (const key in jsonObj) {
if (!jsonObj.hasOwnProperty(key)) {
continue;
}
console.log(key); // Key
console.log(jsonObj[key]); // Value
// Your logic...
}
if you want type safety and don't like decorators
abstract class IPerson{
name?: string;
age?: number;
}
class Person extends IPerson{
constructor({name, age}: IPerson){
super();
this.name = name;
this.age = age;
}
}
const json = {name: "ali", age: 80};
const person = new Person(json);
or this which I prefer
class Person {
constructor(init?: Partial<Person>){
Object.assign(this, init);
}
name?: string;
age?: number;
}
const json = {name: "ali", age: 80};
const person = new Person(json);
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()
}
The best I found for this purpose is the class-transformer
That's how you use it:
Some class:
export class Foo {
name: string;
#Type(() => Bar)
bar: Bar;
public someFunction = (test: string): boolean => {
...
}
}
// the docs say "import [this shim] in a global place, like app.ts"
import 'reflect-metadata';
// import this function where you need to use it
import { plainToClass } from 'class-transformer';
export class SomeService {
anyFunction() {
u = plainToClass(Foo, JSONobj);
}
}
If you use the #Type decorator nested properties will be created, too.
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
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
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);
keeps your classes simple
injection available to the factories for flexibility
I personally prefer option #3 of #Ingo Bürk.
And I improved his codes to support an array of complex data and Array of primitive data.
interface IDeserializable {
getTypes(): Object;
}
class Utility {
static deserializeJson<T>(jsonObj: object, classType: any): T {
let instanceObj = new classType();
let types: IDeserializable;
if (instanceObj && instanceObj.getTypes) {
types = instanceObj.getTypes();
}
for (var prop in jsonObj) {
if (!(prop in instanceObj)) {
continue;
}
let jsonProp = jsonObj[prop];
if (this.isObject(jsonProp)) {
instanceObj[prop] =
types && types[prop]
? this.deserializeJson(jsonProp, types[prop])
: jsonProp;
} else if (this.isArray(jsonProp)) {
instanceObj[prop] = [];
for (let index = 0; index < jsonProp.length; index++) {
const elem = jsonProp[index];
if (this.isObject(elem) && types && types[prop]) {
instanceObj[prop].push(this.deserializeJson(elem, types[prop]));
} else {
instanceObj[prop].push(elem);
}
}
} else {
instanceObj[prop] = jsonProp;
}
}
return instanceObj;
}
//#region ### get types ###
/**
* check type of value be string
* #param {*} value
*/
static isString(value: any) {
return typeof value === "string" || value instanceof String;
}
/**
* check type of value be array
* #param {*} value
*/
static isNumber(value: any) {
return typeof value === "number" && isFinite(value);
}
/**
* check type of value be array
* #param {*} value
*/
static isArray(value: any) {
return value && typeof value === "object" && value.constructor === Array;
}
/**
* check type of value be object
* #param {*} value
*/
static isObject(value: any) {
return value && typeof value === "object" && value.constructor === Object;
}
/**
* check type of value be boolean
* #param {*} value
*/
static isBoolean(value: any) {
return typeof value === "boolean";
}
//#endregion
}
// #region ### Models ###
class Hotel implements IDeserializable {
id: number = 0;
name: string = "";
address: string = "";
city: City = new City(); // complex data
roomTypes: Array<RoomType> = []; // array of complex data
facilities: Array<string> = []; // array of primitive data
// getter example
get nameAndAddress() {
return `${this.name} ${this.address}`;
}
// function example
checkRoom() {
return true;
}
// this function will be use for getting run-time type information
getTypes() {
return {
city: City,
roomTypes: RoomType
};
}
}
class RoomType implements IDeserializable {
id: number = 0;
name: string = "";
roomPrices: Array<RoomPrice> = [];
// getter example
get totalPrice() {
return this.roomPrices.map(x => x.price).reduce((a, b) => a + b, 0);
}
getTypes() {
return {
roomPrices: RoomPrice
};
}
}
class RoomPrice {
price: number = 0;
date: string = "";
}
class City {
id: number = 0;
name: string = "";
}
// #endregion
// #region ### test code ###
var jsonObj = {
id: 1,
name: "hotel1",
address: "address1",
city: {
id: 1,
name: "city1"
},
roomTypes: [
{
id: 1,
name: "single",
roomPrices: [
{
price: 1000,
date: "2020-02-20"
},
{
price: 1500,
date: "2020-02-21"
}
]
},
{
id: 2,
name: "double",
roomPrices: [
{
price: 2000,
date: "2020-02-20"
},
{
price: 2500,
date: "2020-02-21"
}
]
}
],
facilities: ["facility1", "facility2"]
};
var hotelInstance = Utility.deserializeJson<Hotel>(jsonObj, Hotel);
console.log(hotelInstance.city.name);
console.log(hotelInstance.nameAndAddress); // getter
console.log(hotelInstance.checkRoom()); // function
console.log(hotelInstance.roomTypes[0].totalPrice); // getter
// #endregion
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!!!
you can do like below
export interface Instance {
id?:string;
name?:string;
type:string;
}
and
var instance: Instance = <Instance>({
id: null,
name: '',
type: ''
});
My approach is slightly different. I do not copy properties into new instances, I just change the prototype of existing POJOs (may not work well on older browsers). Each class is responsible for providing a SetPrototypes method to set the prototoypes of any child objects, which in turn provide their own SetPrototypes methods.
(I also use a _Type property to get the class name of unknown objects but that can be ignored here)
class ParentClass
{
public ID?: Guid;
public Child?: ChildClass;
public ListOfChildren?: ChildClass[];
/**
* Set the prototypes of all objects in the graph.
* Used for recursive prototype assignment on a graph via ObjectUtils.SetPrototypeOf.
* #param pojo Plain object received from API/JSON to be given the class prototype.
*/
private static SetPrototypes(pojo: ParentClass): void
{
ObjectUtils.SetPrototypeOf(pojo.Child, ChildClass);
ObjectUtils.SetPrototypeOfAll(pojo.ListOfChildren, ChildClass);
}
}
class ChildClass
{
public ID?: Guid;
public GrandChild?: GrandChildClass;
/**
* Set the prototypes of all objects in the graph.
* Used for recursive prototype assignment on a graph via ObjectUtils.SetPrototypeOf.
* #param pojo Plain object received from API/JSON to be given the class prototype.
*/
private static SetPrototypes(pojo: ChildClass): void
{
ObjectUtils.SetPrototypeOf(pojo.GrandChild, GrandChildClass);
}
}
Here is ObjectUtils.ts:
/**
* ClassType lets us specify arguments as class variables.
* (where ClassType == window[ClassName])
*/
type ClassType = { new(...args: any[]): any; };
/**
* The name of a class as opposed to the class itself.
* (where ClassType == window[ClassName])
*/
type ClassName = string & {};
abstract class ObjectUtils
{
/**
* Set the prototype of an object to the specified class.
*
* Does nothing if source or type are null.
* Throws an exception if type is not a known class type.
*
* If type has the SetPrototypes method then that is called on the source
* to perform recursive prototype assignment on an object graph.
*
* SetPrototypes is declared private on types because it should only be called
* by this method. It does not (and must not) set the prototype of the object
* itself - only the protoypes of child properties, otherwise it would cause a
* loop. Thus a public method would be misleading and not useful on its own.
*
* https://stackoverflow.com/questions/9959727/proto-vs-prototype-in-javascript
*/
public static SetPrototypeOf(source: any, type: ClassType | ClassName): any
{
let classType = (typeof type === "string") ? window[type] : type;
if (!source || !classType)
{
return source;
}
// Guard/contract utility
ExGuard.IsValid(classType.prototype, "type", <any>type);
if ((<any>Object).setPrototypeOf)
{
(<any>Object).setPrototypeOf(source, classType.prototype);
}
else if (source.__proto__)
{
source.__proto__ = classType.prototype.__proto__;
}
if (typeof classType["SetPrototypes"] === "function")
{
classType["SetPrototypes"](source);
}
return source;
}
/**
* Set the prototype of a list of objects to the specified class.
*
* Throws an exception if type is not a known class type.
*/
public static SetPrototypeOfAll(source: any[], type: ClassType): void
{
if (!source)
{
return;
}
for (var i = 0; i < source.length; i++)
{
this.SetPrototypeOf(source[i], type);
}
}
}
Usage:
let pojo = SomePlainOldJavascriptObjectReceivedViaAjax;
let parentObject = ObjectUtils.SetPrototypeOf(pojo, ParentClass);
// parentObject is now a proper ParentClass instance
**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;
}
}