如何结合3个firebase observables而不会丢失动态更新 [英] How to combine 3 firebase observables without losing dynamic updates

问题描述

我有一个3平行的firebase要求。我需要把它们作为一个单一的对象。代码如下：

 从@ angular / http导入{Http}; 
从angularfire2导入{AngularFireDatabase}; 
从../model/thread中导入{Thread}; 
从../model/message中导入{Message}; 
从../model/participant导入{Participant}; 
 
构造函数（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 message：Message [] = data [1]; 
让参与者：Participant [] = data [2]; 
让AllUserData = {
线程：线程，
消息：消息，
参与者：参与者
} 
返回AllUserData; 
））
} 
  

我需要的确切格式是这个接口：

pre $ export interface AllUserData {
参与者：Participant [];
线程：Thread [];
消息：消息[];



$ b $问题是forkJoin不会运行.first（）所有的firebase观测。如果我使用.first（）来完成所有的firebase observables，我失去了动态更新，为什么我使用firebase的第一位。我怎样才能得到两全其美的呢？ - 保持firebase的可观察性，但可以加入他们像我的界面格式？我正在把头发拉出来。任何帮助将不胜感激！

解决方案

您可以使用 .forkJoin（...） .combineLatest（.. 。 。

我也简化了 map
$ b

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




I have a 3 parallel firebase requires. And I need to join them as 1 single object. Code as below:

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;
  })
  }

This work and return the exact format I need, which is with this interface:

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

The problem is forkJoin won't run without .first() all the firebase observables. And if I use .first() to complete all the firebase observables, I LOST the dynamic updates which why I use firebase the first place. How can I get the best of both world? - Keeping the firebase observables live but can join them like my interface format? I am pulling my hair out on this. Any help will be greatly appreciated!

解决方案

Instead of .forkJoin(...) you could use .combineLatest(...).

I also did streamline your map a bit:

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 })
)

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