make two properties required if one of them exist | TypeScript Community | Page 1

covert oasis Feb 28, 2024, 9:46 PM

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it may be a big mandela effect, but afaik this should do it

export type Dependency =
    | {
          groupId: string
          artifactId: string
      }
    | {
          repoId: string
          repoUrl: string
          groupId: string
          artifactId: string
      }

meaning, {groupId: '', artifactId: ''} should be valid, {repoId: '', repoUrl: '', groupId: '', artifactId: ''} should be valid too, but {repoId: '', groupId: '', artifactId: ''} should not be valid, bc the repoUrl is missing, and viceversa. but for some reason, this is valid, no errors

const x: Dependency = {repoUrl: '', groupId: '', artifactId: ''}

why? is that normal? i already restarted the TS lsp, it depends on some setting in tsconfig.json? if not, how can i achieve that then?

arctic tiger Feb 28, 2024, 9:48 PM

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Yes it's normal, because TS does not care about excess properties.

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So { repoId: '', groupId: '', artifactId: '' } is treated as { groupId: string, artifactId: string } variant with an excess property of repoId.

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What you probably want is a DU.

covert oasis Feb 28, 2024, 9:49 PM

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since when? is that configurable in any way?

arctic tiger Feb 28, 2024, 9:50 PM

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It's always been like this, it's how TS works.

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!hb discriminated

dry hillBOT Feb 28, 2024, 9:50 PM

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Get Started / TypeScript for Functional Programmers

Discriminated Unions

arctic tiger Feb 28, 2024, 9:50 PM

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But yeah DU is probably what you want ^

covert oasis Feb 28, 2024, 9:51 PM

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so i need to add another property to differentiate them?

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i dont like that behavior at all

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is that normal in other languages too?

arctic tiger Feb 28, 2024, 9:52 PM

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Depends on the language, a lot of languages don't even support union type.

covert oasis Feb 28, 2024, 9:53 PM

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yea

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but

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thats just so weird

arctic tiger Feb 28, 2024, 9:53 PM

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But DU is the proper way to do what you want in TS.

#make two properties required if one of them exist