Question

我正在尝试进行3个连续且异步（无并行）的请求，这些请求带有 HttpClient 和 Observable.forkJoin 。

到目前为止，这是它的外观，首先是 fileOperations 数组：

requestPayloadObjects:any[];//array of objects sent to the same request 

//... after the array was built
this.requestPayloadObjects.forEach( ( payloadObject:any,i:number )=> {
   //_uploadApi.postDoc returns an http observable
   fileOperations.push( this._uploadApi.postDoc( payloadObject ) );
});

然后，当我有一系列可观察的对象后，我使用 Observable.forkJoin

// parallel subscriptions, all 3 observables are subscribed to asyncly
Observable.forkJoin(fileOperations).subscribe((res:any)=> {
        // If all http calls succeeded you get here
    },(err:any)=> {
        // If 1 of all 3 that were sent failed you get here
    });

但是我想要的是：

// NONE-parallel subscription  
Observable.forkJoin(fileOperations).subscribe((res:any)=> {
        // If all http calls succeeded you get here
    },(err:any)=> {
        // If 1 failed, you get here, but you do not send the others
    });

我确定Observable.forkJoin不是实现此目的的Observable方法，而是什么？我如何不进行并行订阅，只有前一个成功，才能继续进行下一个订阅？使用Observables的正确方法是什么？

Answer 1

我认为您正在寻找concatAll。签出rxjs documentation。

Answer 2

按照@MichaelKie的建议使用Observable.concat，我设法通过以下方式解决了此问题。

let requestPayloadObjects:any[] = [{...},{...},{...}];
let firstServerCall:Observable<any>; 

this.requestPayloadObjects.forEach( ( payloadObject:any,i:number )=> {
        let observable = this._uploadApi.postDoc(payloadObject:any);
        if(i===0) {
            firstServerCall = observable ;
        }
        else {
            firstServerCall = firstServerCall.concat(observable );
        };
});
firstServerCall.subscribe((res:any)=> {
  // 1. will get here after every successful response
  // 2. will continue on to send next http call after this
},(err:any)=> {
  // will get here if subscription fails, without continuing to the next observable.
})

角度4顺序异步http请求，错误响应取消

2 个答案: