ES6 Mixin与TypeScript中的泛型类型

2019年7月28日 274次阅读

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

}

但是我不能在BaseClass的派生类上应用MyMixin;与此同时,我也无法制作mixin泛型.

有没有办法使它适用于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() {}
}

最佳答案 我从
TypeScript/PR#13743找到了解决方案,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应该是一个Class(即构造函数),而不是一个类型.

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();
    }
}
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