TS - types - omit field in parameter object in function while keeping types

Question

I have a type that omits specified fields in functions parameter object like this:

type OmitFields<F, K extends string> = F extends (props: infer P) => infer R ? (props: Omit<P, K>) => R : never;

And use it like:

type Omitted = OmitFields<typeof functionWithObjectParam, 'first' | 'second'>

How can I make type parameter K be aware of type that props object gets inferred as, and restrict correct strings to its keys?

Answer 1

I think you want something like:

type OmitFields<
    F extends (props: any) => any,
    K extends keyof Parameters<F>[0],
> = (props: Omit<Parameters<F>[0], K>) => ReturnType<F>

Here F is constrained to be a function that takes one argument. And K is constrained to the keys of that argument.

Then you can reconstruct the function type without any fancy infer stuff.

This seems to do what you want:

function functionWithObjectParam(props: { a: number, b: number }) {}

type Omitted = OmitFields<typeof functionWithObjectParam, 'b'> // ok
type OmittedBad = OmitFields<typeof functionWithObjectParam, 'bad'> // Type '"bad"' does not satisfy the constraint '"b" | "a"'.(2344)

declare const fn: Omitted

fn({ a: 1 }) // ok

fn({ a: 1, b: 2 })
// Argument of type '{ a: number; b: number; }' is not assignable to parameter of type 'Omit<{ a: number; b: number; }, "b">'.
//   Object literal may only specify known properties, and 'b' does not exist in type 'Omit<{ a: number; b: number; }, "b">'.

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