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

1

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 ) {

answered Apr 10, 2022 at 12:07

Elias Soares

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Comments

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

1

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") }

answered Apr 10, 2022 at 12:11

ffuNny

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Comments

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

0

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) {

  }
}

playground

answered Apr 10, 2022 at 12:07

Yoel

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