Make some properties optional in a TypeScript type

Question

If I have a type with all required properties, how can I define another type with the same properties where some of it's properties are still required but the rest are optional?

For example I want a new type derived from SomeType where prop1 and prop2 are required but the rest are optional:

interface SomeType {
    prop1;
    prop2;
    prop3;
    ...
    propn;
}

interface NewType {
    prop1;
    prop2;
    prop3?;
    ...
    propn?;
}

Answer 1

You can use combination of Partial and Pick to make all properties partial and then pick only some that are required:

interface SomeType {
    prop1: string;
    prop2: string;
    prop3: string;
    propn: string;
}

type OptionalExceptFor<T, TRequired extends keyof T> = Partial<T> & Pick<T, TRequired>

type NewType = OptionalExceptFor<SomeType, 'prop1' | 'prop2'>

let o : NewType = {
    prop1: "",
    prop2: ""
}