How to get class name without instance in Typescripts

Question

I'd like to get class name without instance. But it does not work well. Here is my test code.

class User {
    id!:number;
    name!:string
}

//It should use object type. Not used User type directly.
const run = (o:object) => {
    console.log(o.constructor.name);
}

run(User); 
//It printed as Function. 
//But, I'd like to class name(aka, User) without instance(new User)

How can I do?

Does this answer your question? [Get an object's class name at runtime](https://stackoverflow.com/questions/13613524/get-an-objects-class-name-at-runtime) — Jeffrey Ram, Aug 24 '22 at 01:41

Jeffrey Ram · Accepted Answer · 2022-08-24T02:36:00.690

1

Use o.name instead of o.constructor.name

class User {
    id!:number;
    name!:string
}

const run = (o: any) => {
    console.log(o.name);
}

run(User);

Source: Get an object's class name at runtime

edited Aug 24 '22 at 02:36

answered Aug 24 '22 at 01:37

Jeffrey Ram

1,132
1
10
16

When I tested it in my test code(I've linked it in my question), it occur error as "Property 'name' does not exist on type 'object'." – sean s Aug 24 '22 at 02:21
@seans One way to work around it would be to use type `any` instead of `object` like so: `const run = (o:any) => {...}` – Jeffrey Ram Aug 24 '22 at 02:32
If you want it to work for both instances and non-instances, refer to @RahulSharma answer below. I excluded the solution for instances because you said "I'd like to get class name without instance." – Jeffrey Ram Aug 24 '22 at 02:42

score 0 · Answer 2 · answered Aug 24 '22 at 02:27

If o is an object use o.constructor.name otherwise o.name.

class User {
    public name: string;

    constructor(name: string) {
        this.name = name;
    }
}

const run = (o: any) => {
    typeof o === 'object' ? console.log(o.constructor.name) : console.log(o.name);
}

run(User);                     // 'User'
run(new User('john doe'));     // 'User'

How to get class name without instance in Typescripts

2 Answers2