Mongodb聚合嵌套子文档
有文件清单Mongodb聚合嵌套子文档,mongodb,mongodb-query,Mongodb,Mongodb Query,有文件清单 [ { "__v" : 21, "_id" : ObjectId("546330dbb8926d177052e9ff"), "code" : "WfvCc", "description" : "", "elements" : [ { "_id" : ObjectId("546471f61e13b76a0b20ccaf"), "comments" : [],
[
{
"__v" : 21,
"_id" : ObjectId("546330dbb8926d177052e9ff"),
"code" : "WfvCc",
"description" : "",
"elements" : [
{
"_id" : ObjectId("546471f61e13b76a0b20ccaf"),
"comments" : [],
"meta" : {
"createdBy" : "545ab39ef1b0c88a695fcf8d",
"modifiedAt" : "1415868918045",
"createdAt" : "1415868918045"
},
"title" : "awesome title",
"votes" : {
"count" : 3,
"meta" : [
{
"createdBy" : "545ab39ef1b0c88a695fcf8d",
"_id" : ObjectId("546473831e13b76a0b20ccb7"),
"createdAt" : "1415869315618"
},
{
"createdBy" : "545aaddcf1b0c88a695fcf84",
"_id" : ObjectId("546473d71e13b76a0b20ccbc"),
"createdAt" : "1415869399584"
},
{
"createdBy" : "5461c0e2c9c39a192c44226c",
"_id" : ObjectId("546474041e13b76a0b20ccbe"),
"createdAt" : "1415869444056"
}
]
}
}
]
},
{
"__v" : 21,
"_id" : ObjectId("546330dbb8926d177052e9ff"),
"code" : "WfvCc",
"description" : "",
"elements" : [
{
"_id" : ObjectId("546471f61e13b76a0b20ccaf"),
"comments" : [],
"meta" : {
"createdBy" : "545ab39ef1b0c88a695fcf8d",
"modifiedAt" : "1415868918045",
"createdAt" : "1415868918045"
},
"title" : "awesome title",
"votes" : {
"count" : 3,
"meta" : [
{
"createdBy" : "545ab39ef1b0c88a695fcf8d",
"_id" : ObjectId("546473831e13b76a0b20ccb7"),
"createdAt" : "1415869315618"
},
{
"createdBy" : "545aaddcf1b0c88a695fcf84",
"_id" : ObjectId("546473d71e13b76a0b20ccbc"),
"createdAt" : "1415869399584"
},
{
"createdBy" : "5461c0e2c9c39a192c44226c",
"_id" : ObjectId("546474041e13b76a0b20ccbe"),
"createdAt" : "1415869444056"
}
]
}
}
]
}
]
我想汇总用户列表<代码>元素.投票.元.创建数据库,并计算文档中出现的总次数*请注意,elements.vows.meta.createdBy
在每个文档中都是唯一的,因此从理论上讲,这应该使它更简单
到目前为止,我提出了一个问题:
db.sessions.aggregate(
{ $project: {
meta: "$elements.votes.meta"
}},
{ $unwind: "$meta" },
{ $group: {
_id: "voters",
voters: {
$addToSet: "$meta.createdBy"
}
}}
)
只是为了再次完全陷入困境。我知道我需要一个双重分组,只是似乎无法解决。谢谢你的帮助 首先,您应该获得每个“用户”的总数。(即,{$group:{{u id:'$user',count:{'$sum':1}}}) 然后只需按null分组,创建一个包含结果的文档,将每个用户添加到集合中,并将第一次分组的结果推送到数组字段中。(第二组) 结果:
{
"result" : [
{
"_id" : null,
"users" : [
"545ab39ef1b0c88a695fcf8d",
"545aaddcf1b0c88a695fcf84",
"5461c0e2c9c39a192c44226c"
],
"occurances" : [
{
"user" : "5461c0e2c9c39a192c44226c",
"count" : 2
},
{
"user" : "545aaddcf1b0c88a695fcf84",
"count" : 2
},
{
"user" : "545ab39ef1b0c88a695fcf8d",
"count" : 2
}
]
}
],
"ok" : 1
}
真棒,稍加修改,这正是我想要的。关键是将集合加倍
$unwind
。干杯
{
"result" : [
{
"_id" : null,
"users" : [
"545ab39ef1b0c88a695fcf8d",
"545aaddcf1b0c88a695fcf84",
"5461c0e2c9c39a192c44226c"
],
"occurances" : [
{
"user" : "5461c0e2c9c39a192c44226c",
"count" : 2
},
{
"user" : "545aaddcf1b0c88a695fcf84",
"count" : 2
},
{
"user" : "545ab39ef1b0c88a695fcf8d",
"count" : 2
}
]
}
],
"ok" : 1
}