This is my interface
interface X {
key: string
value: number | undefined
default?: number
}
But I want the non-optional keys only, aka. "key" | "value", or just "key" (both will do fine for me)
type KeyOfX = keyof X gives me "key" | "value" | "default".
type NonOptionalX = {
[P in keyof X]-?: X[P]
}
type NonOptionalKeyOfX = keyof NonOptionalX gives "key" | "value" | "default" as -? only removes the optional modifier and make all of them non-optional.
ps. I use Typescript 2.9.
You can use conditional type operator of the form undefined extends k ? never : k to substitute never for keys of values that undefined could be assigned to, and then use the fact that union T | never is just T for any type:
interface X {
key: string
value: number | undefined
default?: number
}
type NonOptionalKeys<T> = { [k in keyof T]-?: undefined extends T[k] ? never : k }[keyof T];
type Z = NonOptionalKeys<X>; // just 'key'
Also this comment might be relevant: https://github.com/Microsoft/TypeScript/issues/12215#issuecomment-307871458
But I want the non-optional keys only, aka. "key" | "value", or just "key" (both will do fine for me)
While easier option of "key" is covered, I'd like to cover a less obvious option of "key" | "value". It might be needed because value is not an optional field. It can be assigned undefined, but it should pass a "value" in x check.
type NonOptionalKeys<T> = { [K in keyof T]-?: T extends { [K1 in K]: any } ? K : never}[keyof T]
How it works:
K) are actually not only types, but types with "requiredness" and "readonlyness" attached;T with -? makes optional keys required before putting them into K;{ [K1 in K]: any } object is an object with single required keynever;never as value;[keyof T] a union of all the value types gets computed, never types get removed ((A | never) = A), and only a union of all required keys is left.Update: readonly is way trickier. Read here or here. You can completely disassemble an object into a tuple like this.
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