How can I get properties of a type as an array in Typescript?

Question

Given the following type:

class Foo {
    constructor(
        private one: string, 
        private two: string, 
        private three: string) {
    }
}

How can I have an array whose values are the type's properties?

eg I need to be able to get an array as:

['One', 'Two', 'Three']

Note I need to have the properties extracted from a type and not an instance otherwise I could simply use Object.keys(instanceOfFoo) .

Answer 1

You can use Reflect.construct() to get the keys, then use Object.keys() to convert that to an array.

Note: if the key doesn't have a default it won't be generated as you can see with four .

 class Foo { constructor() { this.one = '' this.two = '' this.three = '' this.four } } console.log(Object.keys(Reflect.construct(Foo, [])))

Answer 2

Scan the Object.keys(Foo.prototype)

Answer 3

If you just one to list the field's name from a class you can use the following:

let fields = Object.keys(myObj) as Array<MyClass>;

Hope it helps.

Answer 4

You need to instantiate const a = new Foo(); and access using Object.keys

class Foo {
    constructor(
        private one: string, 
        private two: string, 
        private three: string) {
    }
}

const a = new Foo();
console.log(Object.keys(a)); // ["prop"]

DEMO

EDIT: If you want to get from the type it is not possible since types won't be available once the code is compiled.

Answer 5

Since the code above will be transpiled to following js code:

var Foo = (function () {
    function Foo(one, two, three) {
        this.one = one;
        this.two = two;
        this.three = three;
    }
    return Foo;
}());

I'd first extract the constructor using const c = Foo.prototype.constructor and the get the name of arguments from that. This thread shows how to do that.