Typescript 是否可以有 Select 地继承另一个接口

发布于08月11日

是否可以有 Select 地从另一个接口继承

请参见下面的示例

interface A { 
id: string
name: string
gender:string
address: string []
}

interface B { 
id: string
name: string
gender:string
address: number []
}

唯一的区别是属性address，一个是字符串[]，另一个是数字[]

我能这么做吗？

interface B extend A { 
address: number []
}

推荐答案

不，您不能，您将得到一个错误:

Interface 'B' incorrectly extends interface 'A'.
  Types of property 'address' are incompatible.
    Type 'number[]' is not assignable to type 'string[]'.
      Type 'number' is not assignable to type 'string'.

但是您可以做的是使用Omit继承除address之外的所有字段

TS Playground个

interface A { 
    id: string;
    name: string;
    gender: string;
    address: string[];
}

// This doesn't work
/*
interface B extends A { 
    address: number[];
}
*/

// This does
interface B2 extends Omit<A,"address"> {
    address: number[];
}

const myInstance: B2 = {
    id: "1",
    name: "Foo",
    gender: "male",
    address: [1, 2, 3]
};

另一个有效的解决方案是拥有一个通用接口.

interface X<T = number | string> { 
    id: string;
    name: string;
    gender: string;
    address: T[];
}

interface A extends X<string> { 
}

interface B extends X<number> { 
}

const instance: B = {
    id: "1",
    name: "Foo",
    gender: "male",
    address: [1, 2, 3]
};