Question

我的架构： -

var playlistSchema = new Schema({
name        : {type:String,require:true},
videos      : {type:[mongoose.Schema.Types.ObjectId],ref: 'Video'},
},{collection:'playlist'})

var Playlist = mongoose.model('Playlist',playlistSchema);

我在数据库中有一些数据作为例子： -

{
"_id" : ObjectId("58d373ce66fe2d0898e724fc"),
"name" : "playlist1",
"videos" : [ 
    ObjectId("58d2189762a1b8117401e3e2"), 
    ObjectId("58d217e491089a1164a2441f"), 
    ObjectId("58d2191062a1b8117401e3e4"), 
    ObjectId("58d217e491089a1164a24421")
],
"__v" : 0

}

视频的架构是： -

var videoSchema = new Schema({
name        : {type:String,required:true},
createdAt   : {type:String,default:new Date()},
isDisabled  : {type:Boolean,default:false},
album       : {type: mongoose.Schema.Types.ObjectId, ref: 'Album'}
},{collection:'video'})

var Video = mongoose.model('Video',videoSchema);

现在，为了获取播放列表中所有视频的名称，我正在尝试代码： -

 var playlistModel = mongoose.model('Playlist');
let searchParam = {};
searchParam._id = req.params.pid;
playlistModel.findOne(searchParam)
.populate('[videos]')
.exec(function(err,found){
    if(err)
        throw err;
    else{
        console.log(found.videos[0].name);
    }
})

但是我在这里得到了未定义的结果。我没有得到我错的地方plzz任何人帮我解决这个问题。

Answer 1

得到了答案： - 只需更改架构

var playlistSchema = new Schema({
name        : {type:String,require:true},
videos      : [{type:mongoose.Schema.Types.ObjectId,ref: 'Video'}],
},{collection:'playlist'})

var Playlist = mongoose.model('Playlist',playlistSchema);

然后使用

.populate('videos')

而不是

.populate('[videos]')

填充对象ID数组

1 个答案: