How to choose "one of" in a type?

Hi there!

export const user_find_one_controller_input = type({
  "_id?": userApi.schema.user.get("_id"),
  "_email?": userApi.schema.user.get("email")
});

export const user_find_one_controller_input = type({
  "_id?": userApi.schema.user.get("_id"),
  "_email?": userApi.schema.user.get("email")
});

For example here I am wanting to have either or as an object. So saying that if there is no _id then there has to be an "email" and visa versa. Not sure how to achieve this Thanks!

15 Replies

ssalbdivad•2mo ago

Can you use a union like type({ id: "string" }).or({ email: "string" })?

TheGuyOP•2mo ago

Yeah not a bad shout. Wasn't sure if there was a different way to do it. But that should work

ssalbdivad•2mo ago

Yeah, that is the best way to do it in a structural type system like TS/ArkType. Initially the idea of some builtin conditional property syntax is intuitive, but really it's just a different and often more convoluted way to write a union

TheGuyOP•2mo ago

Agreed thanks for quick input One issue may have with union:

ParseError: An unordered union of a type including a morph and a type with overlapping input is indeterminate:
Left: { _id: {} | (In: string) => Out<unknown> }
Right: { email: string }

ParseError: An unordered union of a type including a morph and a type with overlapping input is indeterminate:
Left: { _id: {} | (In: string) => Out<unknown> }
Right: { email: string }

type({_id: objectId}).or({email: "string"})

type({_id: objectId}).or({email: "string"})

Cause I do a custom Arktype that does some pipe stuff. The error makes sense what it is saying though. Just not 100% sure best approach.

ssalbdivad•2mo ago

Are the properties ever both specified or always one or the other?

TheGuyOP•2mo ago

In this case its one or the other. Its like a query that you can choose what you want to search with

ssalbdivad•2mo ago

{ _id: "string", "_email?": "never"} | { "_id?": "never", _email: "string" } Technically you only need to add never to one or the other, but if you want it enforced that way on both

TheGuyOP•2mo ago

Does it matter that technically _id is not actually a string but a morphed type?

ssalbdivad•2mo ago

No that will fix it Because it makes the inputs not overlapping- we will know the intersection of the input of the morph and never is unsatisfiable

TheGuyOP•2mo ago

Yeah thats got it Cool makes sense

ssalbdivad•2mo ago

If in the future you have some very complex relationships you want to check in a case like this, you can always fall back to using .narrow after the base shape has been valdiated

TheGuyOP•2mo ago

Sweet will keep that in mind.

ssalbdivad•2mo ago

Generally best to avoid making your unions to complex when you can check the base shape and then just write whatever logic you want to check those properties

TheGuyOP•2mo ago

Yeah in this case it was fairly straight forward but good to know either way

ssalbdivad•2mo ago

But for a case like this, definitely adding a single ? never is cleaner

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How to choose "one of" in a type?

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