Question

class BaseDoc {
    static Collection;  
    static find(){
        // which is expected to return an instance of SubDoc when run SubDoc.find()
        return this.Collection.find();  
    }
}

class SubDoc extends BaseDoc {}

我希望是：运行SubDoc.find（）时，应用程序知道返回值的类型是SubDoc的实例，而不是BaseDoc的实例。

我该如何实现？

Answer 1

您可以创建通用版本的find，该通用版本将把this参数推断为所调用的类的类型，并使用InstanceType<T>提取实际的实例类型从类中输入：

class BaseDoc {
    static Collection: any;
    static find<T extends typeof BaseDoc>(this: T): InstanceType<T> {
        // which is expected to return an instance of SubDoc when run SubDoc.find()
        return this.Collection.find();
    }
}

class SubDoc extends BaseDoc {

}

SubDoc.find() // return SubDoc

或者您可以强制派生类定义适当类型的通用Collection成员：

class Collection<T> {
    find(): T {
        return null as any // dummy imeplmentation
    }
}
type GetCollectionItem<T extends Collection<any>> = T extends Collection<infer U> ? U: never;

class BaseDoc {
    static Collection: Collection<BaseDoc>;
    static find<T extends { new (... args: any[]) : any, Collection: Collection<any> }>(this: T):  GetCollectionItem<T['Collection']> {
        // which is expected to return an instance of SubDoc when run SubDoc.find()
        return this.Collection.find();
    }
}

class SubDoc extends BaseDoc {
    static Collection: Collection<SubDoc>;
}

SubDoc.find() // return SubDoc

Typescript：如何从继承的静态属性中返回子类类型？

1 个答案: