Is there a way to achieve the following:
type A = { a: string; b: number; c: boolean };
type B = ["b", "a"];
type C = MapProps<A, B> ?? // [number, string]
The solution I have currently is:
type C = {[key in B[number]]: A[key]}
However, this results in an object whereas my goal is to eliminate the keys and retain only the values as a tuple, maintaining the same order as B. Is there a concept similar to Object.values() at the type level?
To solve this problem, we will implement recursive conditional types. The process is outlined as follows:
During each iteration, we extract the first element (K) from the B array and save A[K] in a result array (R). This procedure continues until the B array is empty, at which point we return the final result array.
Here is the implementation code:
type MapProps<
T,
Keys extends PropertyKey[],
Result extends unknown[] = []
> = Keys extends [
infer First extends keyof T,
...infer Rest extends PropertyKey[]
]
? MapProps<T, Rest, [...Result, T[First]]>
: Result;
The type PropertyKey encompasses any key type such as string, number, or symbol.
We utilized the infer keyword to separate the first element and remaining elements in the array type.
For testing purposes, consider the following example:
type A = { a: string; b: number; c: boolean };
type B = ["b", "a"];
type C = MapProps<A, B>; // [number, string]
Try it out on the TypeScript playground
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