CODEWITHSUNDEEP
typescripttypescript-generics
Obi1
Obi1

Reputation: 38

Set only specific properties as required

Typescript has Required<T> utility type which constructs type consisting of all properties of T set to required. But I want to make only specific properties as required, not all. So I've written my own template(sorry about my C++ lexicon) to do it.

interface MyProps {
    foo1?: number;
    foo2?: number;
}

type RequiredProperties<T, RequiredProps extends keyof T> = Required<Pick<T, RequiredProps>> &
  Omit<T, RequiredProps>;

// remove "optional" modifier from the next properties
type requiredProps = 'foo1';
type MakeFoo1Required<T> = RequiredProperties<T, requiredProps>; // <-- got error here

type NewMyProps1 = RequiredProperties<MyProps, requiredProps>; // <-- yes, it works fine
type NewMyProps2 = MakeFoo1Required<MyProps>; // <-- but I wanna to create it through the alias

It works fine, but I got a tsc error

Type '"foo1"' does not satisfy the constraint 'keyof T'.

It happens because tsc does not know that "foo1" is a keyof T.

So I wonder is there a way to solve it?

Upvotes: 0

Views: 96

Answers (1)

Aleksey L.
Aleksey L.

Reputation: 37948

You can add constraint requiring T to have requiredProps:

type MakeFoo1Required<T extends { [K in requiredProps]?: any }> = RequiredProperties<T, requiredProps>;

Playground

Upvotes: 1

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