Mongoose - 如何返回两种不同事物的计数？

Question

我对 MongoDB 和 Mongoose 非常陌生，似乎发现自己需要一个相当复杂的查询。我的架构如下。

const gameSchema = new mongoose.Schema({
    _winnerId: {
        type: mongoose.Schema.Types.ObjectId,
        required: true,
        ref: 'User',
    },
    _loserId: {
        type: mongoose.Schema.Types.ObjectId,
        required: true,
        ref: 'User',
    },
    winnerWords: [String],
    loserWords: [String],
}); 

我正在尝试获取给定用户 ID 的赢/输比率。不知何故，我需要计算用户 ID 作为 _winnerId 出现的次数以及用户 ID 作为 _loserId 出现的次数。这可以通过单个查询来完成吗？

Answer 1

只使用两个同时查询可能会更高效，因为它的计算非常简单，而且赢家/输家字段很容易被索引。但是，我很好奇这是如何实现的。可以这样做：https://mongoplayground.net/p/JbBCzyln9o6

db.collection.aggregate([
  // Combine the user _ids
  {$group: {
      _id: ["$_loserId", "$_winnerId"],
      // Track the amount of times user X lost to user Y
      times: {$sum: 1}
  }},
  
  // Seperate the user _ids
  // Track whether the user won by setting field 'lostWon': 0 if they lost 1 if they won
  {$unwind: {
      path: "$_id",
      includeArrayIndex: "lostWon"
  }},

  // Create a group for each user _id, user the 'lostWon' field to sum the totals
  {$group: {
    _id: "$_id",
    lost: {
      $sum: {
        $multiply: [
          "$times",
          {$cond: [{$eq: ["$lostWon", 0]}, 1, 0]}  // Important: swaps the values
        ]
      }
    },
    won: {
      $sum: {
        $multiply: [
          "$times",
          {$cond: [{$eq: ["$lostWon", 1]}, 1, 0]}  // This line is reduntant, but make query more understandable
        ]
      }
    },
  }},

  // Divide the ratio, if $lost is 0 use the won field value to avoid infinity
  {$addFields: {
    ratio: {
      $cond: [
        {$eq: ["$lost",0]},
        "$won",
        {$divide: ["$won", "$lost"]}
      ]
    }
  }}
])