是否可以使用 mappend/条件类型来转换这个 DU
type MyDU =
| {kind: 'foo'}
| {kind: 'bar'}
type Transformed = DUTransformer<MyDU>
这样我们得到如下结果
type Transformed =
| {kind: 'foo', foo: boolean}
| {kind: 'bar', bar: boolean}
最佳答案
是的,因为 TypeScript 会 distribute mapped types over a union :
type MyDU =
| {kind: 'foo'}
| {kind: 'bar'}
type Kinded<T extends string> = { kind: T }
type DUTransformer<T> = T extends Kinded<infer K> ? T & {[K1 in K]: boolean} : never
type Transformed = DUTransformer<MyDU>
Transformed
的类型是:
type Transformed = ({
kind: 'foo';
} & {
foo: boolean;
}) | ({
kind: 'bar';
} & {
bar: boolean;
})
关于typescript - 扩展受歧视的工会,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64120135/