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Conditional properties through discriminated unions and intersections in TypeScript #8952

Open theguriev opened 1 year ago

theguriev commented 1 year ago

What problem does this feature solve?

https://github.com/vuejs/core/issues/7553 Reopening this because I believe the problem is not solved. It is still impossible to use coditional props.

Just try this after npm create vue@latest inside HelloWorld component

<script setup lang="ts">
interface CommonProps {
  size?: 'xl' | 'l' | 'm' | 's' | 'xs'
}

type ConditionalProps =
  | {
      color?: 'normal' | 'primary' | 'secondary'
      appearance?: 'normal' | 'outline' | 'text'
    }
  | {
      color: 'white'
      appearance: 'outline'
    }

type Props = CommonProps & ConditionalProps
defineProps<Props>()
</script>

and then try to use it

image

Theoretically it should not allow us to use both. That is, there can't be color="white" appearance="text" only color="white" appearance="outline"

What does the proposed API look like?

{
      color?: 'normal' | 'primary' | 'secondary'
      appearance?: 'normal' | 'outline' | 'text'
    }
  | {
      color: 'white'
      appearance: 'outline'
    }
manniL commented 1 year ago

Added another simple example with fully "exclusive" props that should be covered as well and does not work (yet).

goulashify commented 1 year ago

I ran into this many times, huge +1.

DaniilIsupov commented 1 year ago

Typescript allows me to use Conditional Types, but Vue doesn't allow me to do this

louiss0 commented 11 months ago

C'mon Vue and TS NEED FULL SYNERGY.

ThejanNim commented 7 months ago

Hi guys, Any update here? @sxzz if you can please follow up on this issue.

johnsoncodehk commented 5 months ago

This is a type restriction of defineComponent, as a current solution you can use generic. This will bypass defineComponent and define the component as a functional component.

<script setup lang="ts" generic="T">
interface CommonProps {
  size?: 'xl' | 'l' | 'm' | 's' | 'xs'
}
// ...
</script>
occitaneUbald commented 4 months ago

This is a type restriction of defineComponent, as a current solution you can use generic. This will bypass defineComponent and define the component as a functional component.

<script setup lang="ts" generic="T">
interface CommonProps {
  size?: 'xl' | 'l' | 'm' | 's' | 'xs'
}
// ...
</script>

Unfortunately vue-test-utils does not support generic components with Typescript

manniL commented 4 months ago

This is a type restriction of defineComponent, as a current solution you can use generic. This will bypass defineComponent and define the component as a functional component.

<script setup lang="ts" generic="T">
interface CommonProps {
  size?: 'xl' | 'l' | 'm' | 's' | 'xs'
}
// ...
</script>

@johnsoncodehk while props with a discriminator work now in the playground (e.g. color in the example above), two different props as in my comment are still not possible as far as I see.

AdrianFahrbach commented 2 months ago

Any updates on the plans for this? I guess with recent PR #10801 this is in the works?

Sengulair commented 2 days ago

@johnsoncodehk while props with a discriminator work now in the playground (e.g. color in the example above), two different props as in my comment are still not possible as far as I see.↳

It's mostly because of how Typescript is designed due to its structural type system and how Typescript does checks with unions. So in your example, if we simplify it for Typescript, it would be smth like this:

type Props = { one: string } | { other: number };

const Component = (props: Props) => {};

Component({ one: '123', other: 1 }) // no errors

So when you pass some object, it will be checked for each union member separately if it satisfies them. The current object satisfies both of them, so there are no errors.

type Props = { one: string } | { other: number };
const obj = {
  one: 'sad',
  other: 1
} satisfies Props; // no error

So it's mostly not a problem of vue or its typings, it's how Typescript works with unions.

The solution here would be to use discriminated unions (have the same property with different values in union type) or you can manually create a type that will exclude other properties by assigning a never type with an optional mark to fields listed in another union member.

type PropsWithNever = { one: string; other?: never } | { one?: never; other: number };

const objNever1 = { one: 'asd' } satisfies PropsWithNever;
const objNever2 = { other: 1 } satisfies PropsWithNever;

const objNever3 = {
  one: 'asd',
  other: 1
} satisfies PropsWithNever; // shows error

To simplify usage of that I usually use these utilities:

type Without<T, U> = { [P in Exclude<keyof T, keyof U>]?: never };

export type XOR<T, U> = T | U extends object
  ? (Without<T, U> & U) | (Without<U, T> & T)
  : T | U;

type PropsWithXOR = XOR<{ one: string }, { other: number }>;

const objWithXOR = {
  one: 'sad',
  other: 1
} satisfies PropsWithXOR; // shows error

You can check and play with example here

I usually use XOR when I have common props and I can make & operation with XOR result of some exclusive objects. If you also want, you can create XOR utility type for array, not only for two members.

P.S. BTW, I've found nice detailed explanation on this problem in typescript with solution: video