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Mongodb聚合嵌套子文档

转载 作者:行者123 更新时间:2023-12-02 18:06:56 25 4
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[
{
"__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"
}
]
}
}
]
}
]

我想聚合用户列表。跨文档的 elements.votes.meta.createdBy 并计算整个文档中出现的总数。 *请注意,elements.votes.meta.createdBy 每个文档都是唯一的,因此理论上这应该更简单。

到目前为止,我已经得到了一个查询:

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。我知道我需要双重分组,但似乎无法弄清楚。任何帮助表示赞赏。

最佳答案

首先,您应该获得每个“用户”的总数。 (即 { $group: {_id: '$user', count: {'$sum': 1} }})

然后只需按 null 分组即可创建包含结果的文档,添加每个用户进行设置并将第一个分组的结果推送到数组字段。 (第二组)

db.test5.aggregate(
{ $unwind: "$elements" },
{ $unwind: "$elements.votes.meta" },
{ $project: {_id: '$_id', user: '$elements.votes.meta.createdBy'} },
{ $group: {_id: '$user', count: {'$sum': 1} }},
{ $group: {
_id: null,
users: {$addToSet: '$_id'},
occurances: {$push: {'user': '$_id', count: '$count'}}
}
}
)

结果:

{
"result" : [
{
"_id" : null,
"users" : [
"545ab39ef1b0c88a695fcf8d",
"545aaddcf1b0c88a695fcf84",
"5461c0e2c9c39a192c44226c"
],
"occurances" : [
{
"user" : "5461c0e2c9c39a192c44226c",
"count" : 2
},
{
"user" : "545aaddcf1b0c88a695fcf84",
"count" : 2
},
{
"user" : "545ab39ef1b0c88a695fcf8d",
"count" : 2
}
]
}
],
"ok" : 1
}

关于Mongodb聚合嵌套子文档,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28387603/

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