Is it possible to get a union type with all type values from an interface in typescript?
For example, when an interface is given as
interface A {
a: string;
b: () => void;
c: number;
d: string;
e: 'something';
}
the result should be
type B = string | () => void | number | 'something';
I have no clue, how I would approach this problem, if it is even possible.
You can use keyof e.g.
type B = A[keyof A] // will be string | number | (() => void)
The type something will not appear since the compiler does not differentiate between the type string and something - since both are strings it will omit it in the type B.
keyof A) into the key brackets of an interface. Do you know, whether that is mentioned somewhere in the official docs?
keyof returns the keys of A. Secondly, A['a'] will return the type of key a. So it maps over the keys to get their types.A["a" | "b"] = A["a"] | A["b"] (which is basically a basic form of your answer) holds true and still does not seem to be mentioned anywhere in the typescript docs. And I think it's strange, that there does exist an unspoken rule in a language! What if there are other ones? How do I know about them? So my question was actually: Where did you find out about that rule?You need something that is advance typescript.
Below example keyof A is completely interchangeable with string | () => void | number | 'something';. The difference is that if you add another property to A, say myfunc: string, then keyof A will automatically update to be string | () => void | number | something |myfunc. And you can use keyof in generic contexts like pluck, where you can’t possibly know the property names ahead of time. That means the compiler will check that you pass the right set of property names to pluck:
pluck(A, ['number ', 'unknown']); // error, 'unknown' is not in string | () => void | number | 'something';
It can be
let B :keyof A;
