typescript typing with static member of class

Question

I want to access the static member of a class through an object of this class.

I do it using obj.constructor. It works well except with the typescript linter which says

Property 'getName' does not exist on type 'Function'

class Foo {
    public static getName(): string {
        return 'foo';
    }
}

const foo = new Foo();

const name: string = foo.constructor.getName();

I tried using : const name: string = (foo.constructor as Foo).getName();

but it gives me

Property 'getName' is a static member of type 'Foo'

EDIT :

It worked using : const name: string = (foo.constructor as typeof Foo).getName();

Is there any way it could work without manually casting the class?

INFO : I cannot call it directly using Foo.getName() in my specific case

Karol Majewski · Accepted Answer · 2019-09-30 12:37:27Z

4

Static methods are accessed by calling the constructor directly:

Foo.getName()

If for some reason you don't have access to the constructor, assert foo.constructor to be of its type.

const name: string = (foo.constructor as typeof Foo).getName();

edited Sep 30, 2019 at 12:37

answered Sep 30, 2019 at 12:32

Karol Majewski

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4 Comments

Orelsanpls Over a year ago

Thank for your input but I specifically asked to not call it directly using the class name. I would like a solution to work along with foo.constructor.

Karol Majewski Over a year ago

I've seen your hint just after I clicked submit. I edited my answer.

Jared Smith Over a year ago

The problem with this is that if the OP has access to Foo to feed to typeof then he can just call Foo.getName()

Orelsanpls Over a year ago

@JaredSmith totally got it. If the answer is: it's not possible. Alright, but post the answer that way please.

Jared Smith · Accepted Answer · 2019-09-30 12:48:26Z

1

You have to fudge it with the compiler if you don't have access to the class. Note that this is a filthy dirty hack, but sometimes you just have to tell the compiler "shut up, I know what I'm doing".

class Foo {
  public static getName() {
    return 'foo';
  }
}

const foo = new Foo();

interface FudgeIt {
  getName: () => string,
}

// compiler won't let us cast a function to
// a random interface without declaring it
// unknown first.
const c: unknown = foo.constructor;
(c as FudgeIt).getName();

Here's a link to the relevant playground

answered Sep 30, 2019 at 12:48

Jared Smith

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3 Comments

Orelsanpls Over a year ago

I see, so there is no way to implicitly say typescript that .constructor is of typeof Foo alright? Nor any other way to access the static method of an object (different than .constructor) ? If that so, I think i'll just go for the // @ts-ignore for now. Thx for your trick, it may help someone else

Jared Smith Over a year ago

@GrégoryNEUT part of the problem with having an extremely flexible base language like Javascript is that the typechecker has to be very pessimistic about what can or cannot be changed out from under it. You can use the trick or not as you will but if you end up using it I'd include some explanatory comments.

Orelsanpls Over a year ago

I totally aggree. We are moving all our code from javascript to typescript, so we are encountering theses kind of issues. Thx for your help. We will refactor it later to be able to have the type accessible (Foo) and use it properly.

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