使用外部数组展平嵌套的 Observables，没有内部订阅 (RxJS) [英] Flatten nested Observables with outer array without inner subscribe (RxJS)

问题描述

给定以下 TypeScript 函数，它们执行异步 HTTP 调用:

Given the following TypeScript functions, that do async HTTP-Calls:

public retrieveAllMembersIdsFromGroup(groupId: string): Observable<string[]>
public retrieveMember(memberId: string): Observable<Member>

如何将其组合在一个函数中以获取所有成员(Observable)?

How to combine this in one function to get all members (Observable)?

public retieveAllMembersFromGroup(groupId: string): Observable<Member[]>

例如我想要这样的东西:

For example I want to have something like this:

public retieveAllMembersFromGroup(groupId: string): Observable<Member[]> {
return this.retrieveAllMembersIdsFromGroup(groupId).pipe(
      map((membersIds: string[]) => {
        //some magic here to call this.retrieveMember(memberId)
        return //Observable<Member[]>;
      })
    );
}

如果可能，我不想订阅retrieveAllMembersIdsFromGroup，以免必须手动处理id.最好的解决方案是什么?

If possible I do not want to subscribe to retrieveAllMembersIdsFromGroup, so as not to have to process the ids manually. What is the best solution?

推荐答案

以下解决方案对我有用:

Follwoing solution works for me:

public retrieveAllMembersFromGroup(groupId: string): Observable<Member[]> {
    return from(this.retrieveAllMemberIdsFromGroup(groupId)).pipe(
      mergeMap((membersIds: string[]) => {
        const members$: Observable<Member>[] = [];
        membersIds.forEach((memberId: string) => {
          members$.push(this.retrieveMember(memberId));
        });
        return combineLatest(members$);
      })
    );
  }

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