如何在打字稿中获取 T 的 colculated keyof 的类型作为通用类型
how to get type of a colculated keyof T as a generic type in typescript
我有这两个接口
interface PersonRequirements{
user:string,
password:string,
id:number
}
export interface Requirement<R> {
name: keyof R & string,
save: () => any,/* I want this return type to be same as return type of founded key in R*/
}
这是我在其他地方的用例
const idRequirement:Requirement<PersonRequirements>={
name:"id",
save:function ():number/* I want this return type to be same as id's return type(number) but in a generic type safe way*/{
//
}
}
我想使 save()
return 类型与 id 的 return 类型相同,但以通用类型安全的方式我该怎么做?
您可以在编译时声明另一个采用属性名称的通用参数。
export interface Requirement<R, N extends keyof R & string> {
name: N; // this will force the name property to be the same as being passed in
save(): R[N];
}
然后这样使用
const idRequirement: Requirement<PersonRequirements, "id"> ={
name: "id",
save: () => 0
}
我有这两个接口
interface PersonRequirements{
user:string,
password:string,
id:number
}
export interface Requirement<R> {
name: keyof R & string,
save: () => any,/* I want this return type to be same as return type of founded key in R*/
}
这是我在其他地方的用例
const idRequirement:Requirement<PersonRequirements>={
name:"id",
save:function ():number/* I want this return type to be same as id's return type(number) but in a generic type safe way*/{
//
}
}
我想使 save()
return 类型与 id 的 return 类型相同,但以通用类型安全的方式我该怎么做?
您可以在编译时声明另一个采用属性名称的通用参数。
export interface Requirement<R, N extends keyof R & string> {
name: N; // this will force the name property to be the same as being passed in
save(): R[N];
}
然后这样使用
const idRequirement: Requirement<PersonRequirements, "id"> ={
name: "id",
save: () => 0
}