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forkjoin subscribe(rxjs 6)的角度6返回结果

rxjs typescript angular

1

Stéphane GRILLON · 技术社区 · 6 年前

简单案例和确定:

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);
});

控制台是 1 2 3 4 5

我和Ko的更复杂案例:

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);
}

代码执行此操作:

return this.dashboardService.getFoo(foot);

在此之前:

console.log(dataGroup);

如何等待订阅结束,并在重新运行之前(在末尾)在foot.param中添加/mofify所有数据组ID?

1 回复 | 直到 6 年前

1

2

Nima Hakimi 6 年前

这个 todo 函数本身应该是异步的,因此它应该返回 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);
    });
}