可观察中的RXJS可观察

时间:2021-06-13 11:33:08

标签: angular typescript rxjs rxjs-observables

我做了这个代码:

return this.projectService.oneById(id).pipe(mergeMap(project => {
      if (!project) {
        return [];
      }

      const stories = this.getStories(id);

      return combineLatest(project.members.map(member => {
        return this.userService.one(member.id).pipe(map(memberData => {
          const assigned = stories.pipe(mergeMap(t => combineLatest(t.filter(task => {
            if (task && task.assignee?.id === member.id) {
              return {
                ...task,
                id: task.id
              };
            }
          }))));

          return {
            id: member.id,
            name: memberData?.displayName ?? 'Unknown',
            assigned
          };
        }));
      }));
    }));

但是我遇到了一个问题,因为我的函数希望返回 Observable<Type[]>,但它目前正在返回 Observable<{ ...etc, object: Observable<Type[]> }

显然问题是observable里面的observable。但是,我不确定在这种情况下如何解决此问题。我在当前的代码中已经解决了很多次这个问题,但是这个对我来说很难理解,因为无论我尝试什么都没有改变。

感谢您提前提供帮助。

PS:确切的警告是这样的:

Type 'Observable<{ id: string; name: string; assigned: Observable<[UserStory | undefined]>; }[]>' is not assignable to type 'Observable<Member[]>'.   Type '{ id: string; name: string; assigned: Observable<[UserStory | undefined]>; }[]' is not assignable to type 'Member[]'.     Type '{ id: string; name: string; assigned: Observable<[UserStory | undefined]>; }' is not assignable to type 'Member'.       Types of property 'assigned' are incompatible.         Type 'Observable<[UserStory | undefined]>' is missing the following properties from type 'UserStory[]': length, pop, push, concat, and 25 more.

1 个答案:

答案 0 :(得分:1)

threre 是一个地方,你试图将一个 observable 而不是数组作为一个对象作为燃料。这段代码应该更好

return this.projectService.oneById(id).pipe(mergeMap(project => {
  if (!project) {
    return [];
  }
  const stories = this.getStories(id);

  return combineLatest(project.members.map(member => {
    return this.userService.one(member.id).pipe(map(memberData => {
      const assigned$ = stories.pipe(mergeMap(t => combineLatest(t.filter(task => {
        if (task && task.assignee?.id === member.id) {
          return {
            ...task,
            id: task.id
          };
        }
      }))));
      return assigned$.pipe(map(assigned => ({
       id: member.id,
       name: memberData?.displayName ?? 'Unknown',
       assigned
      })));
    }));
  }));
}));