Question

这是我的例子。

items: [
  {
   mart_code : 1,
   director: "James",
   category: "fruit",
   name: "apple",
   money: 5000
  },
  {
   mart_code: 1
   director: "James",
   category: "toy",
   name: "dragon",
   money: 15000
  },
  {
   mart_code: 2,
   director: "Sam",
   category: "fruit",
   name: "orange",
   money: 3500
  }
]

我想得到结果。

martList: [ {
  mart_code: 1,
  directorList:[
  {
   director: "James",
   money: 20000
  }
  ],
  categoryList: [
  {
   category: "fruit",
   money: 5000
  },
  {
   category: "toy",
   money: 15000
  }] 
},
{
  mart_code: 2,
  directorList:[
  {
   director: "Sam",
   money: 3500
  }
  ],
  categoryList: [
  {
   category: "fruit",
   money: 3500
  }]
}
]

我试图$ facet和$ group，$ unwind $ project ... 有没有有效的方法？

Answer 1

希望 $ group 和 $ project 是获得结果的有效方法

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</script>

Answer 2

我尝试过这个。

db.collection.aggregate([
{$unwind : "$items" },
{$group : {
    _id : {
        mart_code : '$mart_code',
        director : '$director'
        },
    money: {$sum : '$money'},
    entry: {$push: {$$ROOT}}
}},
{$group : {
    _id : '$_id.mart_code',
    directorList : {$push : {
        director: "$_id.director",
        money: '$money'
    }},
    entry: {$push: "$entry"}
}},
{ $unwind : "$entry" },
{ $unwind : "$entry" },
{$group : {
    _id : {
        mart_code : '$_id',
        directorList : '$directorList',
        category : '$entry.category'
    },
    money: {$sum: "$entry.money"}
}},
{$group: {
    _id: {
        mart_code : '$_id',
        directorList : '$directorList',
    },
    categoryList:{ $push:{
      category: "$_id.category",
      money: "$money"
    }}
}}
{$project : {
    _id : 0,
    categoryList: "$categoryList",
    directorList: "$_id.directoryList"
}}
])

这是没有效率的方法.....

聚合结果的有效方法

2 个答案: