Typescript - 从另一个联合类型的字段值中获取联合类型

Question

在 Typescript 中，是否可以从另一个联合类型的字段值中获取一个联合类型？

type MyUnionType = 
  | { foo: 'a', bar: 1 }
  | { foo: 'b', bar: 2 } 
  | { foo: 'c', bar: 3 }

// can I generate this automatically? 
// i.e. a union of the possible values of foo in MyUnionType?
type Foos = 'a' | 'b' | 'c'

我希望 Pick<MyUnionType, 'foo'>可能会这样做，但它不太有效 - 它返回我想要的类型但嵌套在 foo 下字段:{ foo: 'a' | 'b' | 'c' }

Answer 1

类型 Foos = MyUnionType['foo'] works只要每个类型都有一个 foo 字段:

type MyUnionType = 
  | { foo: 'a', bar: 1 }
  | { foo: 'b', bar: 2 } 
  | { foo: 'c', bar: 3 }

type FooType = MyUnionType['foo']
// FooType = "a" | "b" | "c"

如果您需要通过异构联合类型进行分发，您可以 filter向下到联合字段中的那些类型:

type PickField<T, K extends string> = T extends Record<K, any> ? T[K] : never;

然后你可以use :

type MyUnionType = 
  | { foo: 'a', bar: 1 }
  | { foo: 'b', bar: 2 } 
  | { foo: 'c', bar: 3 }
  | { bar: 4 }

type FooType = MyUnionType['foo']
// Property 'foo' does not exist on type 'MyUnionType'. :-(

type FooType2 = PickField<MyUnionType, 'foo'>
// type FooType2 = "a" | "b" | "c"

type PickField<T, K extends string> = T extends Record<K, any> ? T[K] : never;
// Alternatively, you could return `undefined` instead of `never`
// in the false branch to capture the potentially missing data