Javascript 按天和项目总数分组，但将项目名称作为关键字输出

我一直在尝试这些例子：和

样本文件：

{ "_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") }

但我的需要是它们的混合体。在push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 } ]
 }

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

在$addToSet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 } ]
 }

{ "_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？如果是的话，任何指导都会很有帮助

根据您的数据，您需要两个阶段，首先收集每个

“项目”

，然后将这些项目详细信息添加到数组中

根据您现有的MongoDB版本，您将如何处理其余部分。对于MongoDB 3.6（从3.4.7开始），您可以使用以下命令重塑数据：

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" }
  }}
])

或者对于早期版本，您可以简单地对结果进行后期处理。无论如何，所有“聚合”工作都是在最后阶段之前完成的：

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 }),
      {}
    )
  }
))

任何一种方法都会产生相同的预期结果：

{
        "_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
        }
}

---

因此，您可以使用新的聚合功能进行操作，但实际上最终结果只是“重塑”，这通常最好留给客户端处理。

您所说的“客户端处理”是什么意思？您是否建议我需要以编程方式“重塑”，而不是在mongodb聚合中？如果是，哪个更快？我至少收藏了10万份文件。