Question

如何通过ID查找房间并确保房间中有当前玩家？

我的mongodb有一个房间的文档，其中包含播放器，而播放器是用户。

const RoomSchema = new Schema({
  players: [{ type: Schema.Types.ObjectId, ref: "Player" }]
})

const PlayerSchema = new Schema({
  user: { type: Schema.Types.ObjectId, ref: "User" }
})

const UserSchema = new Schema({
  username: { type: String}
})

我想找到ID === roomId的房间，并且房间有一个具有user._id === userId的玩家

到目前为止，我的查询仅按ID查找了一个房间，但我想确保返回的房间中有当前用户作为玩家

RoomModel
  .findOne({_id: roomId})
  .populate({ 
    path: 'players',
    populate: {
      path: 'user',
      model: 'User',
      select: ["_id", "username"]
    }
  })

Answer 1

您可以使用mongodb聚合框架执行此任务。

Playground

const result = await RoomModel.aggregate([
  {
    $match: {
      _id: "1",  // match by room id
    },
  },
  {
    $lookup: {
      from: "players",   // must be physical collection name, check if different
      localField: "players",
      foreignField: "_id",
      as: "players",
    },
  },
  {
    $unwind: "$players",
  },
  {
    $match: {
      "players.user": "100", //match by user id
    },
  },
  {
    $lookup: {
      from: "users",
      localField: "players.user",
      foreignField: "_id",
      as: "user"
    }
  }
]);

if (result.length > 0) {
  console.log("found"); //todo: add your logic when found
} else {
  console.log("not found"); //todo: add your logic when not found
}

这将在用户发现时给出类似结果，您可能需要进行一些转换。

[
  {
    "_id": "1",
    "players": {
      "_id": "10",
      "user": "100"
    },
    "user": [
      {
        "_id": "100",
        "username": "user1"
      }
    ]
  }
]

如何基于嵌套对象过滤mongo文档？

1 个答案: