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/
示例文档:
{ "_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体。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" ] }
我想要的是这样的:
{ "_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,任何指南、方向都会有所帮助。
最佳答案
根据您的数据,您需要两个 $group
阶段,以便首先收集每个 “item”
,然后将这些项目详细信息添加到数组中。
根据您可用的 MongoDB 版本,您可以如何处理其余部分。对于 MongoDB 3.6(来自 3.4.7),您可以使用 $arrayToObject
为了 reshape 数据:
db.collection.aggregate([
{ "$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": "$itemsSold" }
}}
])
或者对于早期版本,您可以简单地对结果进行后处理。无论如何,所有的“聚合”工作都是在最后一个阶段之前完成的:
db.collection.aggregate([
{ "$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 }),
{}
)
}
))
两种方式都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来做事,但实际上最终结果只是“ reshape ”,通常最好留给客户端处理。
关于javascript - 按天和项目总计分组,但输出项目名称作为键,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50058304/