Mongodb 在express中将一个mongoose查询的输出用于另一个mongoose查询的输入（异步/等待）

我正在使用express和mongoose实现一个服务器/db。我有一个工作路线，让所有的游戏涉及到一个球员的playerID。我现在正在尝试实现一个可以使用username而不是playerID的

播放器模式：

const mongoose = require('mongoose');

const PlayerSchema = mongoose.Schema( {
  username: {
    type:String,
    required:true,
    unique:true
  },
  date_registered: {
    type: Date,
    default:Date.now
  }
});

module.exports = mongoose.model('Player', PlayerSchema);

游戏模式：

const mongoose = require('mongoose');

const GameSchema = mongoose.Schema( {
  player_1: {
    type: mongoose.Schema.Types.ObjectId,
    ref: 'Player',
    required: true
  },
  player_2: {
    type: mongoose.Schema.Types.ObjectId,
    ref: 'Player',
    required: true
  },
  status: {
    type:String,
  },
  hero_1: {
    type:String
  },
  hero_2: {
    type:String
  },
  date_registered: {
    type: Date,
    default:Date.now
  }

});

module.exports = mongoose.model('Game', GameSchema);

以下是我必须查询的所有游戏，涉及玩家ID的玩家：

//GET GAMES INVOLVING PLAYER BY PLAYER_ID
router.get('/player/:playerId', async (req, res) => {
  try {
    const games = await Game.find({$or:[{ player_1: req.params.playerId }, { player_2: req.params.playerId}]});
    console.log(games)
    res.json(games);
    // weird cuz doesn't throw error if not found, just returns empty list...
  }
  catch (err) {
    res.json({ message: err });
  }
});

以下概述了我想做的事情，但它不起作用，原因肯定有很多：

我试图首先从username获取userId，然后将其传递到游戏的查询中

//GET ALL GAMES ASSOCIATED WITH PLAYER BY USERNAME
router.get('/username/:username', async (req, res) => {
  try {
    const player = await Player.findOne({username:req.params.username});
    console.log(player);
    const games =  Game.find({ $or:[{ player_1: player._id }, { player_2: player._id }] });
    res.json(games);
  }
  catch (err) {
    res.json({ message: err });
  }
});

我一直在读关于.populate（）、promissions和瀑布的书，但我对这方面还不熟悉，希望能得到一些指导

请尝试以下操作：

//GET ALL GAMES ASSOCIATED WITH PLAYER BY USERNAME
router.get('/username/:username', async (req, res) => {
    try {
        const player = await Player.findOne({ username: req.params.username });
        console.log(player);
        /**
         * .findOne() should return a document or null - if no match found..
         */
        if (player) {
            /**
             * .find() will return empty [] only if it didn't find any matching docs but won't throw an error in successful DB operation
             * (irrespective of whether docs matched or not, if op is successful then there will be no error).
             */
            const games = await Game.find({ $or: [{ player_1: player._id }, { player_2: player._id }] }).lean();
            (games.length) ? res.json(games) : res.json(`No games found for ${player._id}`);
        } else {
            res.json('No player found')
        }
    }
    catch (err) {
        res.json({ message: err });
    }
});

---

非常感谢。这成功了！你能解释一下为什么这样做而我的不这样做吗？@emsha：正如我在回答中所说的，如果mongoDB没有找到匹配的文档，它将在

上返回
null
。因此，基本上我们只是使用
if
块来验证我们的DB结果，而不是检查实际错误，所有实际错误都将被
catch
块捕获，而且我想我忘了在游戏开始前放置
wait
。查找
，您可以包含在代码中。。