type a = {id?:number,phone?:string}|{id?:number,name?:string}
type b = ?a
期待: type b = 'id'|'phone'|'name'

type a = {id?:number,phone?:string}|{id?:number,name?:string}
type b = ?a
期待: type b = 'id'|'phone'|'name'
1 CAze May 26, 2020 a & {name?: string} 就好了 |
3 TarotVoyager May 26, 2020 其实更困惑的是 type a = { id?: number, phone?: string, name?: string } 为什么不能是这个格式…… |
4 horizon May 26, 2020 interface A { id?: string; phone?: string; } interface B { id?: string; name?: string; } type C = keyof (A & B); |
5 horizon May 26, 2020 我看错了,无视。。 |
6 TarotVoyager May 26, 2020 |
8 TarotVoyager May 26, 2020 @ytxbnahn #7 上面回复的 stackoverflow 链接应该能解决这个问题,如果用的 ts 2.8 的话 |