Consider the following code snippet:
interface Foo{
one?: string;
two?: string;
}
type Foo2 = {
[P in keyof Foo]: number;
}
It is expected that the type of Foo2 would be { one: number; two: number; }. However, it appears to retain the optional modifier as { one?: number; two?: number; }.
Is there a way to eliminate the optional modifier when using mapped types?
Eliminate the modifier in Typescript 2.8 explicitly with this code:
type Foo2 = {
[P in keyof Foo]-?: number;
}
You can also utilize the built-in Required type for newer versions.
If working with an older version, try this workaround:
type Helper<T, TNames extends string> = { [P in TNames]: (T & { [name: string]: never })[P] };
type Foo3 = Helper<Foo, keyof Foo>;
An option to consider is using Required<T> in place of the shorthand -?
interface Bar {
first?: string;
second?: string;
}
type Bar2 = {
[x in keyof Required<Bar>]: number;
};
If I were to create my own utility, it would likely be something related to typescript
interface Bar {
first?: string;
second?: string;
}
type NeededWithType<U, X> = {
[Q in keyof U]-?: X;
}
// and then simply implement it
type Bar2 = NeededWithType<Bar, Integer>;
I recently authored an article on medium discussing mapped types which you might find intriguing :)
Ever since the release of TypeScript 2.8, you have the option to utilize the Required utility type.
interface Bar{
first?: string;
second?: string;
}
type Bar2 = Required<Bar>
In this scenario, even if 'first' and 'second' have different types, there is no need to be concerned about that.
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