猫鼬-如何返回两个不同事物的计数？

如何解决猫鼬-如何返回两个不同事物的计数？

我对MongoDB和Mongoose非常陌生，似乎发现自己需要立即进行相当复杂的查询。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,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userid作为_winnerId出现的次数和userid作为_loserId出现的次数。可以通过单个查询完成吗？

解决方法

仅使用两个并发查询可能会更有性能，因为这样的计算非常简单，并且可以轻松索引赢家/输家字段。但是，我很好奇如何实现。可以这样完成：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",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: ["$lost","$won",{$divide: ["$won","$lost"]}
      ]
    }
  }}
])