mongodb-获得按字段分组的总和的最大值 [英] mongodb - get highest value of a sum groupped by a field

问题描述

我想我们有此数据

{ "_id" : 1, "item" : "abc", "quantity" : 2, "category" : 1}

{ "_id" : 2, "item" : "jkl", "quantity" : 1, "category" : 1}

{ "_id" : 3, "item" : "xyz", "quantity" : 5, "category" : 2}

{ "_id" : 4, "item" : "abc", "quantity" : 10, "category" : 2}

{ "_id" : 5, "item" : "xyz", "quantity" : 10, "category" : 3}

{ "_id" : 6, "item" : "cds", "quantity" : 9, "category" : 3}

我要查询一个包含最大数量的类别。
，所以我必须：

i want to make a query that give me the category that contain the biggest quantity. so i have to:

1-按类别分组

2-求和每个类别中的项目数量

2- make the sum of quantities of items in each category

3-在这些总和中获得最高价值。

3- get the highest value among those sums.

请帮助我进行此查询。
预先感谢

Please help me to make this query. Thank you in advance

推荐答案

在 $ sort 中使用 $ limit 以获取最大总数：

Use $sort with $limit to get the maximum total:

db.items.aggregate([
  { $match: {} },
  { $group: {
         _id: '$category',
         sumQ: {$sum: '$quantity'}
      } 
  },
  { $sort: {sumQ: -1 } },
  { $limit: 1 }
])

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