Question

我正在尝试在TypeScript中使用ES6 Mixin。我所拥有的内容如下所示，它与BaseClass完美配合。

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，我无法对来自MyMixin的派生类应用BaseClass;与此同时，我也无法制作mixin泛型。

有没有办法让它适用于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

我找到了来自TypeScript/PR#13743的解决方案，并通过@Maximus的评论对其进行了优化。

这段代码有效。原因是，在MyMixin中，T应该是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();
    }
}

ES6 Mixin与TypeScript中的泛型类型

1 个答案: