firebase - 如何在不丢失动态更新的情况下组合 3 个 firebase observable

Question

我需要 3 个并行 Firebase。我需要将它们作为 1 个单个对象加入。代码如下:

import {Http} from "@angular/http";
import {AngularFireDatabase} from "angularfire2";
import {Thread} from "../model/thread";
import {Message} from "../model/message";
import {Participant} from "../model/participant";

constructor(private  http: Http, private db:AngularFireDatabase) { }

loadUserThreads(uid): Observable<AllUserData> {
return Observable.forkJoin([
  this.db.list('participant/' + uid + '/threads').first(),
  this.db.list('participant/' + uid + '/messages').first(),
  this.db.list('participant/' + uid + '/participants').first()
])
  .map((data:any[]) => {
    let threads: Thread[] = data[0];
    let messages: Message[] = data[1];
    let participants: Participant[] = data[2];
    let AllUserData = {
      threads: threads,
      messages: messages,
      participants: participants
    }
    return AllUserData;
  })
  }

此工作并返回我需要的确切格式，这是使用此接口(interface)的:

export interface AllUserData {
  participants: Participant[];
  threads: Thread[];
  messages: Message[];
}

问题是如果没有 .first() 所有 firebase observables，forkJoin 将无法运行。如果我使用 .first() 来完成所有 firebase observables，我会丢失动态更新，这就是我首先使用 firebase 的原因。我怎样才能两全其美？ - 保持 firebase observables 活跃，但可以像我的界面格式一样加入它们？我对此很抓狂。任何帮助将不胜感激!

Answer 1

您可以使用 .combineLatest(...) 而不是 .forkJoin(...) .

我还简化了你的 map :

Observable.combineLatest(
  this.db.list('participant/' + uid + '/threads'),
  this.db.list('participant/' + uid + '/messages'),
  this.db.list('participant/' + uid + '/participants')
)
.map(([threads, messages, participants]: [Thread[], Message[], Participant[]]) =>
  ({ threads, messages, participants })
)