Get Type keys in TypeScript

Question

Lets say we have this Type:

export type UsersSchema = {
  id: number;
  firstName: string;
  lastName: string;
  email: string;
};

Is there a way to make this pseudo-code work:

Object.keys(UsersSchema) which would ideally yield: ['id', 'firstName', 'lastNight', 'email']

Obviously UsersSchema is not a value, so the above does not work...

Does this answer your question? [Get keys of a Typescript interface as array of strings](https://stackoverflow.com/questions/43909566/get-keys-of-a-typescript-interface-as-array-of-strings) — captain-yossarian from Ukraine, Mar 10 '21 at 13:21
There is no easy way. You can use third party library for it https://stackoverflow.com/questions/43909566/get-keys-of-a-typescript-interface-as-array-of-strings . But, I don't know what about order guarantee — captain-yossarian from Ukraine, Mar 10 '21 at 13:22

spender · Answer 1 · 2021-03-10T13:52:27.037

The type doesn't exist at run time.

However, (leaning heavily on this beautiful answer, which will require TS4.x because of its use of recursive conditional types), you can create a tuple type that enforces a tuple with the required names.

So:

type TupleUnion<U extends string, R extends string[] = []> = {
    [S in U]: Exclude<U, S> extends never 
                ? [...R, S] 
                : TupleUnion<Exclude<U, S>, [...R, S]>;
}[U] & string[];


export type UsersSchema = {
  id: number;
  firstName: string;
  lastName: string;
  email: string;
};

const allKeysOfUsersSchema: TupleUnion<keyof UsersSchema> = 
    ["id", "firstName", "lastName", "email"]; //OK

const wrongKeysOfUsersSchema: TupleUnion<keyof UsersSchema> = 
    ["monkey", "firstName", "lastName", "email"]; //error

const missingKeysOfUsersSchema: TupleUnion<keyof UsersSchema> = 
    ["id", "firstName", "lastName"]; //error

const tooManyKeysOfUsersSchema: TupleUnion<keyof UsersSchema> = 
    ["id", "firstName", "lastName", "email", "cat"]; //error

OK... so you'll have to maintain this separate tuple, manually, but at least if things change, the compiler will push you into remedial action, so type-safety is maintained.

Playground link

score 3 · Answer 2 · answered Mar 10 '21 at 13:24

3

You can use keyof type operator.

export type UsersSchema = {
  id: number;
  firstName: string;
  lastName: string;
  email: string;
};

type Keys = keyof UsersSchema // "id" | "firstName" | "lastName" | "email"

answered Mar 10 '21 at 13:24

hwasurr

31
2

Is there a way to use them as value now? `fetch("https://my-url?filter=id,firstName,lastName,email")`? – Matthis Kohli Mar 21 '22 at 12:57
@MatthisKohli As far as i know, it's impossible to use type as a value. However, you can use this key type to create type-checking functions – hwasurr Mar 25 '22 at 00:16

Pavel Bondarenko · Answer 3 · 2022-06-26T16:57:33.323

3

in addition to hwasurr answer you probably need to put the keys in parenthesis to get the right type

the array of keys:

type Keys = (keyof UsersSchema)[]
// ("id" | "firstName" | "lastName" | "email")[]

edited Jun 26 '22 at 16:57

answered Jun 26 '22 at 16:56

Pavel Bondarenko

31
4

Get Type keys in TypeScript

3 Answers3

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