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 .
Issue :订阅回调一直无限期执行
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();
}
}
我尝试过(同样的问题):
-
在Http.service中返回一个Observable而不是Promise
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回调内部(也使用setTimeout(()=> ....))
1 回答
正如forkJoin reference所说,它
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 .