Get properties of a class using Typescript

Question

Is there a way to get properties names of class in TypeScript: in the example I would like to 'describe' the class A or any class and get an array of its properties (maybe only public one ?), is it possible? Or should I instantiate the object first?

class A {
    private a1;
    private a2;
    /** Getters and Setters */

}

class Describer<E> {
    toBeDescribed:E ;
    describe(): Array<string> {
        /**
         * Do something with 'toBeDescribed'                          
         */
        return ['a1', 'a2']; //<- Example
    }
}

let describer = new Describer<A>();
let x= describer.describe();
/** x should be ['a1', 'a2'] */ 

Answer 1

Just for fun

class A {
    private a1 = void 0;
    private a2 = void 0;
}

class B extends A {
    private a3 = void 0;
    private a4 = void 0;
}

class C extends B {
    private a5 = void 0;
    private a6 = void 0;
}

class Describer {
    private static FRegEx = new RegExp(/(?:this\.)(.+?(?= ))/g); 
    static describe(val: Function, parent = false): string[] {
        var result = [];
        if (parent) {
            var proto = Object.getPrototypeOf(val.prototype);
            if (proto) {
                result = result.concat(this.describe(proto.constructor, parent));
            } 
        }
        result = result.concat(val.toString().match(this.FRegEx) || []);
        return result;
    }
}

console.log(Describer.describe(A)); // ["this.a1", "this.a2"]
console.log(Describer.describe(B)); // ["this.a3", "this.a4"]
console.log(Describer.describe(C, true)); // ["this.a1", ..., "this.a6"]

Update: If you are using custom constructors, this functionality will break.

Answer 2

Another solution, You can just iterate over the object keys like so, you must instantiate the object first:

printTypeNames<T>(obj: T) {
    const objectKeys = Object.keys(obj) as Array<keyof T>;
    for (let key of objectKeys)
    {
       console.Log('key:' + key);
    }
}

Answer 3

Some answers are partially wrong, and some facts in them are partially wrong as well.

Answer your question: Yes! You can.

In Typescript

class A {
    private a1;
    private a2;


}

Generates the following code in Javascript:

var A = /** @class */ (function () {
    function A() {
    }
    return A;
}());

as @Erik_Cupal said, you could just do:

let a = new A();
let array = return Object.getOwnPropertyNames(a);

But this is incomplete. What happens if your class has a custom constructor? You need to do a trick with Typescript because it will not compile. You need to assign as any:

let className:any = A;
let a = new className();// the members will have value undefined

A general solution will be:

class A {
    private a1;
    private a2;
    constructor(a1:number, a2:string){
        this.a1 = a1;
        this.a2 = a2;
    }
}

class Describer{

   describeClass( typeOfClass:any){
       let a = new typeOfClass();
       let array = Object.getOwnPropertyNames(a);
       return array;//you can apply any filter here
   }
}

For better understanding this will reference depending on the context.

Answer 4

This TypeScript code

class A {
    private a1;
    public a2;
}

compiles to this JavaScript code

class A {
}

That's because properties in JavaScript start extisting only after they have some value. You have to assign the properties some value.

class A {
    private a1 = "";
    public a2 = "";
}

it compiles to

class A {
    constructor() {
        this.a1 = "";
        this.a2 = "";
    }
}

Still, you cannot get the properties from mere class (you can get only methods from prototype). You must create an instance. Then you get the properties by calling Object.getOwnPropertyNames().

let a = new A();
let array = return Object.getOwnPropertyNames(a);

array[0] === "a1";
array[1] === "a2";

Applied to your example

class Describer {
    static describe(instance): Array<string> {
        return Object.getOwnPropertyNames(instance);
    }
}

let a = new A();
let x = Describer.describe(a);