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node.js - mongoDB 中的多个 $group

转载 作者:太空宇宙 更新时间:2023-11-03 21:54:36 25 4
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我在 mongodb 中有不同的记录。我写了一个小例子:

{_id:"sad547er4w2v5x85b8", name:"Jhon", jobTime:600, floor:2, dept:5, age:25},
{_id:"xcz547wer4xcvcx1g2", name:"Alex", jobTime:841, floor:4, dept:1, age:55},
{_id:"xcnwep2321954ldfsl", name:"Alice", jobTime:100, floor:3, dept:3, age:55},
{_id:"23s3ih94h548jhfk2u", name:"Anne", jobTime:280, floor:2, dept:8, age:22},
{_id:"03dfsk9342hjwq1503", name:"Alexa", jobTime:355, floor:2, dept:6, age:25}

我尝试获得此输出,但我不知道如何按两次分组以获得该结构。

{[
{age:22, floors:[{floor:2,persons:[{name:"Anne",jobTime:280,dept:8}]}]},
{age:25, floors:[{floor:2,persons:[{name:"Jhon",jobTime:600,dept:5},{name:"Alexa",jobTime:355,dept:6}]}]},
{age:55, floors:[{floor:3,persons:[{name:"Alex",jobTime:841,dept:1}]},{floor:4,persons:[{name:"Alice",jobTime:100,dept:3}]}]}
]}

最佳答案

正是如此。使用“两个”$group阶段

collection.aggregate([
{ "$group": {
"_id": {
"age": "$age",
"floor": "$floor",
},
"persons": { "$push": {
"name": "$name",
"jobTime": "$jobTime",
"dept": "$dept"
}}
}},
{ "$group": {
"_id": "$_id.age",
"floors": { "$push": {
"floor": "$_id.floor",
"persons": "$persons"
}}
}}
],function(err,results) {
// deal with results here
})

产生:

{ 
"_id" : 25,
"floors" : [
{ "floor" : 2,
"persons" : [
{ "name" : "Jhon", "jobTime" : 600, "dept" : 5 },
{ "name" : "Alexa", "jobTime" : 355, "dept" : 6 }
]
}
]
},
{
"_id" : 55,
"floors" : [
{ "floor" : 3,
"persons" : [
{ "name" : "Alice", "jobTime" : 100, "dept" : 3 }
]
},
{ "floor" : 4,
"persons" : [
{ "name" : "Alex", "jobTime" : 841, "dept" : 1 }
]
}
]
},
{
"_id" : 22,
"floors" : [
{ "floor" : 2,
"persons" : [
{ "name" : "Anne", "jobTime" : 280, "dept" : 8 }
]
}
]
}

所以最初的 $group位于复合键上,包括您要添加到初始“数组”中的项目的详细信息(对于“persons”)。然后是第二个$group仅将初始 _id 的一部分作为其键,并再次将内容“插入”新数组中。

关于node.js - mongoDB 中的多个 $group,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44565499/

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