Question

我对Observables和rxjs的整个概念是陌生的，所以这可能很明显。但是，由于我急于学习，请提供帮助

当用户单击以在博客文章下显示评论时，我很高兴

 showComments(){
    
    this.firestoreservice.getComments(this.job.id).subscribe( data =>{
      this.comments = data //type [] of comment objects with (message,posterID,timeposted...)
    })

    this._showcomments = true;

  }

此函数调用一个服务函数，该函数将可观察到的内容返回到该作业的我的注释列表。在可观察的订阅中，初始化一个变量来保存我的所有评论。但是，在执行此操作之前，是否可以用匹配的用户名替换每个评论的posterID属性？参见下面的预期结果：

showComments(){
   
   this.firestoreservice.getComments(this.job.id).subscribe( data =>{

     //for each comment held in data, replace comment.posterID with getUserName(posterID)
     //getUserName(posterID) will also return an observable which emmits one string
     //then append comment to comments list (or add back in to data then destructively assign 
     //comments=data )
     
     this.comments = data //type [] of comment objects with (message,posterID,timeposted...)
   })

   this._showcomments = true;

 }

Answer 1

它看起来有点像这样。...

this.firestoreservice.getComments(this.job.id).pipe(
  switchMap(comments => { // switchMap to subscribe to inner
    const getUsers$ = comments.map(comment => // array map comments into userName fetch
      this.firestoreservice.getUserName(comment.posterID).pipe(
        // once you have the username, rx map back into the comment with the username assigned
        map(userName => ({...comment, posterID: userName})),
        first() // use to complete after one emission to make forkJoin work
      )
    )
    return forkJoin(...getUsers$); // forkJoin them all together for parallel execution
  });
).subscribe(data => {
  this.comments = data // now just set the data
})

在返回外部可观察对象之前，如何修改可观察对象发出的每个对象的内部元素

1 个答案: