MongoDB聚合:如何对对象数组进行分组，获取组的倍数，然后对倍数进行求和

Question

文档

[    
    { _id: "a1", 
      selections:[
          {_id:"s1", questionId:"q1", points:3, group:"no-group"},
          {_id:"s2", questionId:"q2", points:2, group:"group-1"},
          {_id:"s3", questionId:"q3", points:3, group:"no-group"},
          {_id:"s4", questionId:"q4", points:7, group:"group-2"},
          {_id:"s5", questionId:"q5", points:8, group:"group-2"},
          {_id:"s6", questionId:"q6", points:9, group:"group-1"},
      ],
      userId: "u1"
    },
]

我正在尝试创建一个聚合，它可以对对象数组进行分组，获取组的多个，然后对倍数进行求和。例如，对于上面给定的文档，分组数组如下所示：

[           
    [
          {_id:"s1", questionId:"q1", points:3, group:"no-group"},
          {_id:"s3", questionId:"q3", points:3, group:"no-group"},
    ],
    [
          {_id:"s2", questionId:"q2", points:2, group:"group-1"},
          {_id:"s6", questionId:"q6", points:9, group:"group-1"},
    ],
    [
          {_id:"s4", questionId:"q4", points:7, group:"group-2"},
          {_id:"s5", questionId:"q5", points:8, group:"group-2"},
    ],
]

然后得到除group: "no-group"组外的所有组的倍数。用于"no-group"的值与它们之和，而不是得到它们的倍数。

例如，分组数组将导致：

[
    [3+3],// for no-group sum the points
    [2*9],// for group-1 multiply the points
    [7*8],// for group-2 multiply the points
]
// output = [ 6, 18, 56 ]

然后将输出6+18+56=80之和。

我如何在mongodb聚合中做到这一点？这样我就能得到如下的输出

{ userId:"u1", totalPoints:"80" }

Answer 1

这个怎么样？

db.getCollection('col1').aggregate([
    { $unwind: '$selections' }, // unwrap initial selections array
    { $group: { _id: { user: '$userId', group: '$selections.group' }, points: { $push: '$selections.points' } } }, // group by user and group
    { $group: { _id: '$_id.user', totalPoints: { $sum: { // get the total sum for...
        $cond:{
            if: { $eq: ['$_id.group', 'no-group'] },
            then: { $sum: '$points'}, // the sum of points if no-group
            else: { $reduce: { input: '$points', initialValue: 1, in: { $multiply: ['$$value', '$$this']} } } // the multiplication of points for rest of cases
        }
    } } } }
])