我正在做一些http请求并使用rxjs成功通知结果:
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请求都完成后,我想在getReportings方法内部迭代报告数组,读取一些值,并为每个报告再次生成带有这些值的新http请求。
总而言之,我有2个(会员/报告)+ appr。 4到8(其他东西)请求。
当所有appr。完成了6到8个请求我希望从成功处理程序中的前6到8个请求中获取所有数据。
如何使用rxjs执行此操作?
更新
当用户olsn要求提供更多细节时,我理解他现在关注他的问题我在这里添加了更多数据(伪代码)6到8个请求应该是什么样子:
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]
});
}
答案 0 :(得分:4)
您可以使用switchMap
扩展您的信息流,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]
});
}
这应该这样做,它更像是一种“旧式逻辑锤”方法 - 如果你正在寻找它:可能有一种更优雅的方法来解决这个问题其他运营商,但在不了解完整数据和所有逻辑的情况下很难说。