Can I implement such type in typescript?
type Someone = {
who: string;
}
type Someone = Someone.who === "me" ? Someone & Me : Someone;
-
1The example doesn't make sense because a compile time type cannot depend on a runtime value.Ingo Bürk– Ingo Bürk2021-05-30 20:00:16 +00:00Commented May 30, 2021 at 20:00
Add a comment
|
2 Answers
I think a discriminated union could work:
type Me = {
who: 'me',
secret: string;
}
type Other = {
who: 'other',
}
type Someone = Me | Other;
1 Comment
ZhenyaUsenko
damn, that actually works for me and looks rly simple) thanks, just need to rewrote all interfaces to types now because this thing doesn't work with interfaces
Yes, you can do so with the help of Generics, Distributive Conditional Types and Unions.
Below is a minimally working example:
type Someone<T extends string> = {
who: T;
}
type Me = {
me: boolean;
}
type Thisone<T extends string> = T extends 'me' ? Someone<T> & Me : Someone<T>;
function whoami<T extends string>(who: T) {
return {
who,
me: who === 'me' ? true : undefined
} as Thisone<T>
}
const a = whoami('you');
const b = whoami('me');
a.who; // ok
a.me; // error
b.who; // ok
b.me; // ok
3 Comments
ZhenyaUsenko
i think in your example it is required to specify "T" parameter, i would like it to take a value automatically e.g ``` const a: Someone = { who: "me", me: true } // ok const b: Someone = { who: "you", me: true } // error ```
yqlim
@ZhenyaUsenko that is impossible because that would mean depending on a runtime value. Type checking only happens in the compile time, which is why you need to set the type constraint with
T.ZhenyaUsenko
btw, the answer above works for me, thanks for the help anyways