我正在执行一些http请求,并使用rxjs成功地通知结果:
getReportings(departmentId: number): Observable<any> {
return Observable.forkJoin(
this.http.get('/api/members/' + departmentId).map(res => res.json()),
this.http.get('/api/reports/' + departmentId).map(res => res.json())
);
}当两个http请求都完成时,我希望在getReportings方法中迭代reports数组,读取一些值,并为每个报表再次发出一个带有这些值的新http请求。
总之,我有两个(成员/报告)+应用程序。4至8(其他东西)请求。
当所有的人。6到8个请求已经完成,我希望在成功处理程序中从前面的6到8个请求中获取所有数据。
我怎样才能用rxjs做到这一点呢?
更新
由于用户olsn要求提供更多的详细信息,我现在了解了他的担忧,我在这里输入了更多数据(伪代码),6到8个请求应该是什么样的:
getReportings(departmentId: number): Observable<any> {
return Observable.forkJoin(
this.http.get('/api/members/' + departmentId).map(res => res.json()),
this.http.get('/api/reports/' + departmentId).map(res => res.json())
).switchMap((result: [any[], any[]]) => {
let members: any[] = result[0];
let reports: any[] = result[1];
let allNewStreams: Observable<any>[] = [
Observable.of(members),
Observable.of(reports)
];
for(let report of reports)
{
allNewStreams.push(
this.http.get(report.url + ?key1=report.name1?).map(res => res.json()));
}
return Observable.forkJoin(allNewStreams); // will contain members, reports + 4-6 other results in an array [members[], reports[], ...other stuff]
});
}发布于 2016-12-17 23:50:29
可以使用switchMap扩展您的流,如下所示:
getReportings(departmentId: number): Observable<any> {
return Observable.forkJoin(
this.http.get('/api/members/' + departmentId).map(res => res.json()),
this.http.get('/api/reports/' + departmentId).map(res => res.json())
).switchMap((result: [any[], any[]]) => {
let members: any[] = result[0];
let reports: any[] = result[1];
let allNewStreams: Observable<any>[] = [
Observable.of(members),
Observable.of(reports)
];
// do your stuff and push new streams to array...
if (foo) { // for each additional request
let reportId: string | number = reports[0].id; // or however you retrieve the reportId
allNewStreams.push(
this.http.get('some/api/ + bar)
.map(res => res.json())
.map(data => ({reportId, data})); // so your final object will look like this: {reportId: "d38f68989af87d987f8", data: {...}}
);
}
return Observable.forkJoin(allNewStreams); // will contain members, reports + 4-6 other results in an array [members[], reports[], ...other stuff]
});
}这应该可以做到,它更像是一种“旧式-逻辑-锤子”的方法--以防你在寻找它:也许有一种更优雅的方法可以通过使用其他操作符来解决这个问题,但如果不知道完整的数据和所有的逻辑,那就很难说了。
https://stackoverflow.com/questions/41204055
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