是否可以从任何“n”个定义的接口扩展并在 TypeScript 中创建一个新的子接口？

Question

I would like to extend from any of 3 defined interfaces and create a new one with its own properties.

Eg:

interface IChild extends any(IFirst, ISecond, IThird) {
  ownVar1: string;
  ownVar2: number;
}

Is there any way to achieve this.

NB I have used any here as just an example as to show that would like to use some similar keyword if available,

Answer 1

Here is the type you are looking for:

interface A {
    a: 'a'
}
interface B {
    b: 'b'
}
interface C {
    c: 'c'
}

type UnionInterface = A | B | C;

type Child = UnionInterface & {
    ownVar1: string;
    ownVar2: number;
}


const x: Child = {
    ownVar1: 'hello',
    ownVar2: 42,
    a: 'a'
}

const y: Child = {
    ownVar1: 'hello',
    ownVar2: 42,
    b: 'b'
}

const z: Child = {
    ownVar1: 'hello',
    ownVar2: 42,
    c: 'c'
}

Please keep in mind, that in TS you can do smth like that:

interface IChild extends A, B, C {

}

Bit it is mean that your IChild interface extends all properties from A,B,C.

If you want to extend a union type, A or B or C, you should use types instead interfaces

Answer 2

I am not sure what you want to accomplish. Maybe you can add some more information about why. My first thought was that you wanted to use an Interface that is either IFirst , ISecond and so on. Then a function can take any of those as an argument, no matter the type (thus your IChild ). If that is the case, I would instead create an IParent interface and let IFirst , ISecond and IThird extend IParent

interface IParent{}
interface IFirst extends IParent {}
interface ISecond extends IParent {}

// then do something like this
function foo(IParent myInterface){}

const first: IFirst = {};
const second: ISecond = {};

foo(first); // OK
foo(second); // OK