javascript – Angular DI和继承：注入基本服务的扩展

我有一个基本服务和两个继承服务：

@Injectable({ providedIn: 'root' })
export class BaseService {
  foo(src?: string){
    return `speaking from ${src || 'BaseService'}`;
  }
}


@Injectable({ providedIn: 'root' })
export class SomeService extends BaseService {
  foo(){
    return super.foo('SomeService')
  }
}

@Injectable({ providedIn: 'root' })
export class AnotherService extends BaseService {
  foo(){
    return super.foo('AnotherService')
  }
}

我希望将它们注入到某个组件中并检索三个单独类的实例：

@Component({
    selector: 'my-app',
    template: `
        <div>
            <p>Who's there?</p>
            <p>{{ base }}</p>
            <p>{{ some }}</p>
            <p>{{ another }}</p>

        </div>
    `,
})
export class App {
    base: string;
    some: string;
    another: string;

    constructor(base: BaseService, some: SomeService, another: AnotherService) {
        this.base = base.foo();
        this.some = some.foo();
        this.another = another.foo();

    }
}

相反,我得到了同一个类的三个实例(HTML输出)：

Who's there?

speaking from BaseService

speaking from BaseService

speaking from BaseService

>为什么这不起作用？
>为什么SomeService,AnotherService和BaseService不是Angular DI的唯一令牌？

似乎放

...    
{ provide: SomeService , useClass: SomeService },
{ provide: AnotherService , useClass: AnotherService },
...

在提供商将使其工作.

>为什么明确需要这个？

一个plnkr：https://next.plnkr.co/edit/BvmppLHRbFbz9CFZ

解决方法:

SomeService和AnotherService从BaseService继承装饰器元数据,因此angular会在其位置注入BaseService实例.

这很危险,因为调用SomeService或AnotherService中的任何实例成员都不会从BaseService继承,这会触发运行时错误.

存档您正在寻找的行为的最简单方法是从公共抽象基类继承,没有装饰器：

export abstract class AbstractBaseService {
  foo(src?: string) {
    return `speaking from ${src || 'AbstractBaseService'}`;
  }
}

@Injectable({ providedIn: 'root' })
export class BaseService extends AbstractBaseService {
  foo() {
    return super.foo('BaseService');
  }
}

@Injectable({ providedIn: 'root'})
export class SomeService extends AbstractBaseService {
  foo() {
    return super.foo('SomeService');
  }
}

@Injectable({ providedIn: 'root' })
export class AnotherService extends AbstractBaseService {
  foo() {
    return super.foo('AnotherService');
  }
}

我modified your plnkr测试这种方法.