How do I initialize a TypeScript object with a JSON object
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!)
Answer
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);