Typescript: Omit a property from all interfaces in a union, but keep the union structure
You want to distribute the Omit across a union. Luckily, you can use distributive conditional types to achieve this:
type DistributiveOmit<T, K extends keyof any> = T extends any ? Omit<T, K> : never;The T extends any construction looks like it doesn't do much, but since T is a type parameter, it distributes the conditional type across any union constituents of T.
Let's test it:
type CC = DistributiveOmit<C, "toRemove">;// type CC = Pick<A, "key1" | "key2"> | Pick<B, "key1" | "key3">You can verify that this is equivalent to the CC type you want.
Hope that helps; good luck!
Another approach is:
export declare type Without<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>;type CC = Without<A, "toRemove">| Without<B, "toRemove">;