英文:
How to use infer to extract a common key from a union type?
问题
interface Base<T extends string> {
name: T;
}
interface Foo extends Base<'Foo'> {}
interface Bar extends Base<'Bar'> {}
interface Baz extends Base<'Baz'> {}
export type FooBarBaz = Foo | Bar | Baz;
// I want to make a type, which is equivalent to this:
// export type FooBarBazTypes = 'Foo' | 'Bar' | 'Baz';
// But I want to do it with some magic infer stuff.
export type SomeMagicCrap<T> = T extends Base<infer U> ? U : never;
export type FooBarBazTypes = SomeMagicCrap<FooBarBaz> // Should be equivalent to 'Foo' | 'Bar' | 'Baz';
如何提取'Foo'、'Bar'和'Baz'？我知道它涉及到infer，但我不记得具体是怎么做的。
英文:
I'm creating a library where I need to use a common type distinguished by name; I need to be able to get the names easily.
interface Base<T extends string> {
name: T;
}
interface Foo extends Base<'Foo'> {}
interface Bar extends Base<'Bar'> {}
interface Baz extends Base<'Baz'> {}
export type FooBarBaz = Foo | Bar | Baz;
// I want to make a type, which is equivalent to this:
// export type FooBarBazTypes = 'Foo' | 'Bar' | 'Baz';
// But I want to do it with some magic infer stuff.
export type SomeMagicCrap<T>;
export type FooBarBazTypes = SomeMagicCrap<FooBarBaz > // Should be equivalent to 'Foo' | 'Bar' | 'Baz';
How can I extract 'Foo', 'Bar' and 'Baz'? I know it involves infer but I don't remember exactly how.
答案1
得分: 1
你可以检查FooBarBaz联合类型（Foo | Bar | Baz）的值是否扩展了Base，然后通过infer推断传递给Base的泛型值，并将其返回，以生成一个新的联合类型从这些字符串值中。
interface Base<T extends string> {
name: T;
}
interface Foo extends Base<'Foo'> {}
interface Bar extends Base<'Bar'> {}
interface Baz extends Base<'Baz'> {}
export type FooBarBaz = Foo | Bar | Baz;
type SomeMagicCrap<T> = T extends Base<infer U> ? U : never;
export type FooBarBazTypes = SomeMagicCrap<FooBarBaz>; // 'Foo' | 'Bar' | 'Baz'
英文:
What you can do is check whether values of the FooBarBaz union (Foo | Bar | Baz) extend Base and then infer the generic value passed to Base and return it to generate a new union from those string values.
interface Base<T extends string> {
name: T;
}
interface Foo extends Base<'Foo'> {}
interface Bar extends Base<'Bar'> {}
interface Baz extends Base<'Baz'> {}
export type FooBarBaz = Foo | Bar | Baz;
type SomeMagicCrap<T> = T extends Base<infer U> ? U : never;
export type FooBarBazTypes = SomeMagicCrap<FooBarBaz>; // 'Foo' | 'Bar' | 'Baz'
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