typescript construct to handle optional object type

Question

I have a typescript function, findUser that I would like to have as inputs either a username (as a string) or a user_id as a number.

function findUser(info: { user_id: number } | { username: string}) { if (info.user_id) { } if (info.username) { } }

I'm getting the typescript error on the info.user_id clearly because the user_id field it is not guaranteed to exist on info.

I realize an alternative way to do this is to define the function as

function findUser(info: { user_id?: number; username?: string}) {

and then it works, however clearly you can pass in an empty object and it would be nice to have a way to check against this.

(Also, I have a more-complicated structure with 4 types of inputs. And I think if I solve this simpler example, I can determine the more complicated one.)

My question: is there a way in typescript to do something like the first function definition?

Elias Soares · Accepted Answer · 2022-04-10 12:07:50Z

You need to define both alternatives as type:

 type UserInfo = { user_id: number; username?: string; } | { user_id?: number; username: string }; function findUser(info: UserInfo ) {

Note that this way will allow providing both user_id and username. If this is not desired, you can use never type:

 type UserInfo = { user_id: number; username?: never; } | { user_id?: never; username: string }; function findUser(info: UserInfo ) {

ffuNny · Accepted Answer · 2022-04-10 12:11:25Z

you can use interfaces or a fast way like this

const findUser: { (name: string) : void; (Id: number): void; } = (value: string | number): void => { if (typeof value === typeof 'string') { console.log("find by name") }

Yoel · Accepted Answer · 2022-04-10 12:07:08Z

I would do it that way

interface PropsName { userName: string userId?: string } interface PropsID { userId: string userName?: string } function findUser(info: PropsName | PropsID) { if (info.userId) { } if (info.userName) { } }

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