Question

我一直在尝试这些例子：https://docs.mongodb.com/manual/reference/operator/aggregation/push/和 https://docs.mongodb.com/manual/reference/operator/aggregation/addToSet/

示例文件：

{ "_id" : 1, "item" : "abc", "price" : 10, "quantity" : 2, "date" : ISODate("2014-01-01T08:00:00Z") }
{ "_id" : 2, "item" : "jkl", "price" : 20, "quantity" : 1, "date" : ISODate("2014-02-03T09:00:00Z") }
{ "_id" : 3, "item" : "xyz", "price" : 5, "quantity" : 5, "date" : ISODate("2014-02-03T09:05:00Z") }
{ "_id" : 4, "item" : "abc", "price" : 10, "quantity" : 10, "date" : ISODate("2014-02-15T08:00:00Z") }
{ "_id" : 5, "item" : "xyz", "price" : 5, "quantity" : 10, "date" : ISODate("2014-02-15T09:05:00Z") }
{ "_id" : 6, "item" : "xyz", "price" : 5, "quantity" : 5, "date" : ISODate("2014-02-15T12:05:10Z") }
{ "_id" : 7, "item" : "xyz", "price" : 5, "quantity" : 10, "date" : ISODate("2014-02-15T14:12:12Z") }

但我需要的是它们的混合物。在推送示例中，结果如下所示：

{
    "_id" : { "day" : 46, "year" : 2014 },
    "itemsSold" : [
        { "item" : "abc", "quantity" : 10 },
        { "item" : "xyz", "quantity" : 10 },
        { "item" : "xyz", "quantity" : 5 },
        { "item" : "xyz", "quantity" : 10 }
    ]
}
{
    "_id" : { "day" : 34, "year" : 2014 },
    "itemsSold" : [
        { "item" : "jkl", "quantity" : 1 },
        { "item" : "xyz", "quantity" : 5 }
    ]
 }
 {
     "_id" : { "day" : 1, "year" : 2014 },
     "itemsSold" : [ { "item" : "abc", "quantity" : 2 } ]
 }

在$ addToSet示例中，结果如下所示：

{ "_id" : { "day" : 46, "year" : 2014 }, "itemsSold" : [ "xyz", "abc" ] }
{ "_id" : { "day" : 34, "year" : 2014 }, "itemsSold" : [ "xyz", "jkl" ] }
{ "_id" : { "day" : 1, "year" : 2014 }, "itemsSold" : [ "abc" ] }

我想要的是：

{ "_id" : { "day" : 46, "year" : 2014 }, "itemsSold" : { "xyz": 25, "abc": 10 } }
{ "_id" : { "day" : 34, "year" : 2014 }, "itemsSold" : { "xyz": 5, "jkl": 1 ] }
{ "_id" : { "day" : 1, "year" : 2014 }, "itemsSold" : { "abc": 2 } }

这可能吗？如果是，任何指导，方向都会有所帮助。

Answer 1

根据您的数据，您需要两个$group阶段，以便先按"item"收集，然后将这些项目详细信息添加到数组中。

根据您的MongoDB版本，您可以使用的是如何处理其余版本。对于MongoDB 3.6（来自3.4.7），您可以使用$arrayToObject来重塑数据：

db.collection.aggregate([
  { "$group": {
    "_id": {
      "year": { "$year": "$date" },
      "dayOfYear": { "$dayOfYear": "$date" },
      "item": "$item"
    },
    "total": { "$sum": "$quantity" }
  }},
  { "$group": {
    "_id": {
      "year": "$_id.year",
      "dayOfYear": "$_id.dayOfYear"
    },
    "itemsSold": { "$push": { "k": "$_id.item", "v": "$total" } }
  }},
  { "$sort": { "_id": -1 } },
  { "$addFields": { 
    "itemsSold": { "$arrayToObject": "$itemsSold" }
  }}
])

或者使用早期版本，您只需发布处理结果即可。所有＆＃34;聚合＆＃34;无论如何，工作在最后阶段之前完成：

db.collection.aggregate([
  { "$group": {
    "_id": {
      "year": { "$year": "$date" },
      "dayOfYear": { "$dayOfYear": "$date" },
      "item": "$item"
    },
    "total": { "$sum": "$quantity" }
  }},
  { "$group": {
    "_id": {
      "year": "$_id.year",
      "dayOfYear": "$_id.dayOfYear"
    },
    "itemsSold": { "$push": { "k": "$_id.item", "v": "$total" } }
  }},
  { "$sort": { "_id": -1 } },
  /*
  { "$addFields": { 
    "itemsSold": { "$arrayToObject": "$itemsSold" }
  }}
  */
]).map( d => Object.assign( d, 
  { 
    itemsSold: d.itemsSold.reduce((acc,curr) =>
      Object.assign(acc, { [curr.k]: curr.v }),
      {}
    )
  }
))

无论哪种方式产生相同的期望结果：

{
        "_id" : {
                "year" : 2014,
                "dayOfYear" : 46
        },
        "itemsSold" : {
                "xyz" : 25,
                "abc" : 10
        }
}
{
        "_id" : {
                "year" : 2014,
                "dayOfYear" : 34
        },
        "itemsSold" : {
                "jkl" : 1,
                "xyz" : 5
        }
}
{
        "_id" : {
                "year" : 2014,
                "dayOfYear" : 1
        },
        "itemsSold" : {
                "abc" : 2
        }
}

所以你可以用新的聚合功能做事，但实际上最终的结果只是'＃34;重塑＆＃34;通常最好留给客户处理。

按日分组和项目总计，但输出项目名称为键

1 个答案: