1

如何通过 ID 找到房间并确保房间中有当前玩家?

我的 mongodb 有一个房间文档,其中包含玩家,玩家是用户。

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,并且房间有一个带有 user._id === userId 的玩家

到目前为止,我的查询仅通过 ID 找到一个房间,但我想确保返回的房间有当前用户作为玩家

RoomModel
  .findOne({_id: roomId})
  .populate({ 
    path: 'players',
    populate: {
      path: 'user',
      model: 'User',
      select: ["_id", "username"]
    }
  })
4

1 回答 1

2

您可以为此任务使用 mongodb 聚合框架。

操场

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
}

当用户发现时,这将给出这样的结果,您可能需要进行一些转换。

[
  {
    "_id": "1",
    "players": {
      "_id": "10",
      "user": "100"
    },
    "user": [
      {
        "_id": "100",
        "username": "user1"
      }
    ]
  }
]
于 2020-04-05T16:33:20.843 回答