javascript - $.getJSON 只返回部分和一个空数组

Question

我正在创建一个对象来处理 YouTube API，我有两种方法:

• getCommentList - 获取当前上传的 url，例如 http://gdata.youtube.com/feeds/api/videos/VIDEO_ID/comments?alt=json 并返回一个对象数组 - 评论的作者和评论的内容。
• getEntriesObject - 返回一个包含每个上传条目对象的数组，我们有标题、缩略图和从 getCommentList
返回的评论列表

我的 jQuery 代码:

var Youtube = {

   getCommentObject : function(url){
      if( url ){
         var currentCommentFeed = {},
             commentsList = [];

         $.getJSON(url,function(data){
            $.each(data.feed.entry,function(index){
               currentCommentFeed = this;

               commentsList.push({
                  author : currentCommentFeed.author[0].name.$t,
                  content : currentCommentFeed.content.$t
               });

            });

            return commentsList;
         });

      }
   },

   getEntriesObject : function(){
      var username = 'SOMEYOUTUBEUSERHERE',
          url = 'http://gdata.youtube.com/feeds/api/users/' + username + '/uploads?alt=json',
          currentEntry = {},
          currentObject = {},
          entryList = [];

      // Scope fix
      var that = this;

      $.getJSON(url,function(data){
         $.each(data.feed.entry, function(index){

            // Caching our entry
            currentEntry = this;

            // Adding our entry title and thumbnail
            currentObject = {
               title: currentEntry.title.$t
            };

            if(currentEntry.media$group.media$thumbnail.length == 4)
               currentObject['thumbnail'] = currentEntry.media$group.media$thumbnail[3].url;

            // Let`s get the comments - undefined....
            currentObject['comments'] = that.getCommentObject(currentEntry.gd$comments.gd$feedLink.href + "?alt=json");

            console.log(currentObject);
            entryList.push(currentObject);
         });
      });

      return entryList;

   }

   /*

   entry[i].title.$t
   entry[i].gd$comments.gd$feedLink.href + "?alt=json"
   entry[i].media$group.media$thumbnail[3]

   // Comments 
   entry[i].author.name.$t
   entry[i].author.content.$t
   */
};

我有 console.log(currentObject) 并且正在获取标题。但是我没有得到缩略图 URL 和评论。

此外，当我运行 getEntriesObject 时，我得到一个空数组。

Answer 1

当您在 $.getJSON 的回调中调用 return 时，您只返回该回调函数，而不是“外部”getCommentObject。因此，当您稍后调用 that.getCommentObject 时，您不会得到任何返回 (undefined)。

getCommentObject: function(url){
   if( url ){

      // Snip ...

      $.getJSON(url,function(data){

         // Snip ...

         return commentsList; // <- Here
      });

   }
}

要修正这个，让 getCommentObject 接受一个回调函数。

getCommentObject: function(url, callback){
   if( url ){

      // Snip ...

      $.getJSON(url,function(data){

         // Snip

         // Remove the return statement
         callback(commentsList);
      });

   }
}

像这样调用这个函数:

that.getCommentObject(
    currentEntry.gd$comments.gd$feedLink.href + "?alt=json", 
    function (commentsList) { 
       currentObject['comments'] = commentsList; 
});

替换

currentObject['comments'] = that.getCommentObject(currentEntry.gd$comments.gd$feedLink.href + "?alt=json");