Question

简单案例，然后确定：

const observables = [];
for (let i = 0; i < this.originalData.length; i++) {
      observables.push( this.dashboardService.getDetails(this.id1[i], this.id2[i])
       };

forkJoin(...observables).subscribe(dataGroup => {
    console.log(dataGroup.id);
});

控制台为1 2 3 4 5

对我和KO来说，情况更复杂：

private todo(foot) {
    const observables = [];
    for (let i = 0; i < this.originalData.length; i++) {
      observables.push( this.dashboardService.getDetails(this.id1[i], this.id2[i])
    };

    forkJoin(...observables).subscribe(dataGroup => {
      console.log(dataGroup);
      // How to put all dataGroup id in foot.param ?
    });

    return this.dashboardService.getFoo(foot);
}

代码执行以下操作：

return this.dashboardService.getFoo(foot);

在此之前：

console.log(dataGroup);

如何在重新调整（最后）之前等待订阅结束并在foot.param中添加/确定所有dataGroup id？

Answer 1

todo函数本身应该是异步的，因此它应该返回一个Observable：

private todo(foo): Observable<any> {
    const observables = [];
    for (let i = 0; i < this.originalData.length; i++) {
      observables.push( this.dashboardService.getDetails(this.id1[i], this.id2[i])
    };

    // notice that dataGroup is an array of latest value of all observables
   return forkJoin(observables).map((dataGroup: any[]) => {
      // do whatever you want to foo before calling the function
      // remember you need to also merge if getFoo returns an observable as well
      // but I assume it doesn't
      return this.dashboardService.getFoo(foo);
    });
}

forkJoin订阅的angular 6返回结果（rxJs 6）

1 个答案: