MongoDB聚合和嵌套数字数组

Posted 2023-03-11

【中文标题】MongoDB聚合和嵌套数字数组

【英文标题】：MongoDB Aggregate sum nested number array

【发布时间】：2021-08-21 02:06:00

【问题描述】：

我有以下文档结构

            "_id": "60b7b7c784bd6c2a1ca57f29",
            "user": "607c58578bac8c21acfeeae1",
            "exercises": [
                
                    "executed_reps": [8,7],
                    "_id": "60b7b7c784bd6c2a1ca57f2a",
                    "exercise_name": "Push up"
                ,
                
                    "executed_reps": [5,5],
                    "_id": "60b7b7c784bd6c2a1ca57f2b",
                    "exercise_name": "Pull up"
                
            ],
        

在聚合中，我尝试对所有 executed_reps 求和，因此本示例中的最终值应为 25 (8+7+5+5)。

这是我目前的代码：

const exerciseStats = await UserWorkout.aggregate([
    
        $match: 
            user:  $eq: ObjectId(req.query.user) ,
        ,
    ,
     $unwind: '$exercises' ,
    
        $group: 
            _id: null,
            totalReps: 
                $sum: 
                    $reduce: 
                        input: '$exercises.executed_reps',
                        initialValue: '',
                        in:  $add: '$$this' ,
                    ,
                ,
            ,
        ,
    ,
]);

totalReps 的结果为 5。我做错了什么？

【参考方案1】：

10 分钟后我自己找到了解决方案：

UserWorkout.aggregate([
    
        $match: 
            user:  $eq: ObjectId(req.query.user) ,
        ,
    ,
     $unwind: '$exercises' ,
    
        $project: 
            total:  $sum: '$exercises.executed_reps' ,
        ,
    ,
    
        $group: 
            _id: null,
            totalExercises:  $sum: '$total' ,
        ,
    ,])

必须先使用 $project。 :)

【参考方案2】：

你可以这样做：

const exerciseStats = await UserWorkoutaggregate([
  
    "$addFields": 
      "total": 
        "$sum": 
          "$map": 
            "input": "$exercises",
            "as": "exercise",
            "in": 
              "$sum": "$$exercise.executed_reps"
            
          
        
      
    
  
])

下面是工作示例：https://mongoplayground.net/p/5_fsPgSh8EP