Question

const artistSchema = new Schema({
  name: String,
});

const albumSchema = new Schema({
  name: String,
  artists: [{type: ObjectId, ref: 'Artist'}]
});

const trackSchema = new Schema({
  title: String,
  album: {type: ObjectId, ref: 'Album'},
  visits: Number,
});

graphql看起来像这样：

tracks {
  title
  visits
  album {
    name
    artists {
      name
    }
  }
}

如何获得仅具有艺术家ID的访问量最大的曲目？

我尝试使用Track.find({'album.artists.id': artistId})。无法正常工作。

Answer 1

万一有人需要它：

const artistId = "...";
const limit = 5;

const artistObjectId = mongoose.Types.ObjectId(artistId);

return await Track.aggregate([
  {
    $lookup: {
      from: "albums",
      localField: "album",
      foreignField: "_id",
      as: "album"
    }
  },
  {
    $match: {
      "album.artists": artistObjectId
    },
  },
  {
    $sort: {
      visits: -1
    }
  },
  {
    $limit: limit
  },
  {
    $lookup: {
      from: "artists",
      localField: "album.artists",
      foreignField: "_id",
      as: "album.artists"
    }
  },
]);

按嵌套的populate（）字段过滤

1 个答案: