angular – 如何取消订阅forkJoin返回的Observable?

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在我的Angular2-typescript应用程序中,我只使用forkJoin在所有并行HTTP调用完成后返回一个Observable.

问题:订阅回调一直无限期地执行

这是我的代码

http.service

import {Http} from "@angular/http";

constructor (private _http: HTTP) {}

makeGetRequest(....) {
    return this._http.get(URL)
           .map (res => res.json)
           .toPromise();

my.service

import {Observable} from "rxjs/Observable";
import {HttpService} from "http.service"

constructor (private _httpService: HttpService) {}

myMethod(): Observable<any[]> {
 return Observable.forkJoin(
            this._httpService.makeGetRequest(
                URL1
            ),this._httpService.makeGetRequest(
                URL2
            )
        )
}

my.component

import MyService from "my.service";
import Subscription from "rxjs";

constructor (private _service: MyService) {}

mySub: Subscription;

ngOnInit() {
    this.mySub = this._service.myMethod.subscribe(data => {
         data.forEach(console.log(data));
         this.mySub.unsubscribe();
     }
}

我尝试了什么(同样的问题):

>在Http.service而不是Promise中返回一个Observable
>在my.component中使用.first().subscribe()而不是subscribe()
>把this.mySub.unsubscribe();在ngOnInit的末尾而不是在subscribe回调中(也使用setTimeout(()=> ….))

正如 forkJoin reference所说,它

Runs all observable sequences in parallel and collect their last elements.

这意味着运算符从已完成的observable中获取值,并返回具有单个值的已完成的observable.没有必要取消订阅.

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