问题
I have types like this:
type GenericType<T, K extends keyof T = keyof T> = {
name: K;
params: T[K]
}
type Params = {
a: 1;
b: 2;
}
const test: GenericType<Params> = {
name: "a",
params: 2
}
When I create an object like test that has property name: "a" I want the type of params to be inferred so that params must be 1. In my example params has type 1 | 2 which is Params[keyof Params]. But since name is "a" I think it should be possible to limit the type of params to just 1 without specifying the second generic type like const test: GenericType<Params, "a">. Basically what I want is:
type GenericType<T> = {
name: keyof T;
params: T[value of name]
}
Is this possible with typescript?
回答1:
Sure, you want GenericType<T> to be a union, which can be accomplished by mapping over the properties of T and then joining that mapped type into a union via lookup type:
type GenericType<T> = {
[K in keyof T]: {
name: K;
params: T[K]
}
}[keyof T];
Now you'll get the behavior you want, I think:
const test: GenericType<Params> = { // error!
name: "a",
params: 2
}
const good: GenericType<Params> = {
name: "b",
params: 2
}
You can verify that the type GenericType<Params> evaluates to { name: "a"; params: 1; } | { name: "b"; params: 2; }, so you're required to match the correct name with the correct params.
Okay, hope that helps; good luck!
Playground link to code
来源:https://stackoverflow.com/questions/60342467/typescript-infer-type-of-generic-keyof