Typescript union types to keys on other union type

Question

I've got a union type, say:

type BooleanOperator = "and" | "or";

I want to create another union type that is based on it:

type ConditionMap_Good = { and: string[] } | { or: string[] }

However, I've only been able to do this manually, written as above, which won't be feasible when the original union has many cases.

I know of mapped types, but what they generate doesn't match what I want:

type ConditionMap_Bad = {
    [K in BooleanOperator]: string[]
}

// type ConditionMap_Bad = {
//     and: string[];
//     or: string[];
// }

Is there a concept or construct in Typescript to help me define what I want?

Does this answer your question? [TypeScript: Map union type to another union type](https://stackoverflow.com/questions/51691235/typescript-map-union-type-to-another-union-type) — bugs, Sep 09 '22 at 11:51
No, it doesn't, because in that question the initial union type is used as a value in the final union, while I want to use it as a key. — Ionuț-Claudiu Herciu, Sep 09 '22 at 11:53

score 3 · Accepted Answer · answered Sep 09 '22 at 13:02

As the linked answer says, you need to use distributive conditional types to "map" over the union and you will need to use a mapped type to get the string to be key of the object type.

type ConditionMap = 
  BooleanOperator extends infer U extends string
    ? U extends U 
      ? { [K in U]: string[] }
      : never
    : never

Here is also a reusable generic variant.

type ToConditionMap<T extends string> = 
  T extends T
    ? { [K in T]: string[] }
    : never

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Typescript union types to keys on other union type

1 Answers1