ES6 Mixin 与 TypeScript 中的泛型类型

Question

我正在尝试在 TypeScript 中使用 ES6 Mixin。我所拥有的如下所示，它与BaseClass.

class BaseClass {
    public foo() {}
};

interface IMyMixin {
    foo2();
}

let MyMixin = (superclass: typeof BaseClass) => class extends BaseClass implements IMyMixin {
    public foo2() {}
}

class MyBaseClass extends MyMixin(BaseClass) {

}

但是我不能申请MyMixin派生类 from BaseClass; 同时，我也不能使 mixin 通用。

有没有办法让它同时适用于BaseClass和DerivedClass？

class DerivedClass extends BaseClass {
    public bar() {}
}

class MyDerivedClass extends MyMixin(DerivedClass) {
    public something() {
// Compile Error: Property 'bar' does not exist on type 'MyDerivedClass'
        this.bar();
    }
}

// Compile Error: 'T' only refers to a type, but is being used as a value here.
let MyMixin = <T extends BaseClass>(superclass: typeof T) => class extends T implements IMyMixin {
    public foo2() {}
}

Answer 1

我从TypeScript/PR#13743找到了解决方案，并根据@Maximus 的评论对其进行了优化。

这段代码有效。原因是， inMyMixin应该T是一个Class（即构造函数），而不是类型。

type Constructor<T> = new (...args: any[]) => T;

class BaseClass {
    public foo() { }
};

interface IMyMixin {
    foo2();
}

// `implements IMyMixin` is optional.
let MyMixin = <T extends Constructor<BaseClass>>(superclass: T) => class extends superclass implements IMyMixin {
    public foo2() { }
}

class DerivedClass extends BaseClass {
    public bar() {}
}

class MyDerivedClass extends MyMixin(DerivedClass) {
    public something() {
        this.bar();
        this.foo();
        this.foo2();
    }
}