MongoDB - 在使用 $group 进行 $lookup 后回退一个 $unwind 嵌套数组
MongoDB - Rewind an $unwind nested array after $lookup using $group
MongoDB 聚合在一分钟内变得呈指数级复杂化!
我正在 $unwind
一个嵌套数组,然后对展开的嵌套数组中的每个对象的 _id
执行 $lookup
。我最后的尝试是用 $group
反转平仓。但是,我无法重建原始的嵌入式数组,其原始 属性 名称以及每个文档的其余原始直接属性。
这是我目前的尝试:
db.users.aggregate([
{
$unwind: "$profile",
$unwind: {
path: "$profile.universities",
preserveNullAndEmptyArrays: true
}
},
{
$lookup: {
from: "universities",
localField: "profile.universities._id",
foreignField: "_id",
as: "profile.universities"
}
},
{
$group: {
_id: "$_id",
emails: { "$first": "$emails" },
profile: { "$first": "$profile" },
universities: { "$push": "$profile.universities" }
}
}
]).pretty()
我得到的是这样的:
{
"_id" : "A_USER_ID",
"emails" : [
{
"address" : "AN_EMAIL_ADDRESS",
"verified" : false
}
],
"profile" : {
"name" : "NAME",
"company" : "A COMPANY",
"title" : "A TITLE",
"phone" : "123-123-1234",
"disabled" : false,
"universities" : [
{
"_id" : "ID_1",
"name" : "UNIVERSITY_NAME_1",
"code" : "CODE_1",
"styles" : {AN_OBJECT}
}
]
},
"universities" : [
[
{
"_id" : "ID_1",
"name" : "UNIVERSITY_NAME_1",
"code" : "CODE_1",
"styles" : {AN_OBJECT}
}
],
[
{
"_id" : "ID_2",
"name" : "UNIVERSITY_NAME_2",
"code" : "CODE_2",
"styles" : {AN_OBJECT}
}
]
]
}
此结果有 2 个问题:
- 结果
universities
是一个数组数组,每个数组包含一个对象,因为 $lookup
为原始 $profile.universities
嵌套数组返回了单个元素数组。它应该只是一个对象数组。
- 生成的
universities
应该位于其嵌套在 profiles
下的原始位置。我知道为什么原来的 profile.universities
是这样的,因为我使用的是 $first
运算符。我这样做的目的是保留 profile
的所有原始属性,同时保留原始嵌套的 universities
数组。
最终,我需要的是这样的:
{
"_id" : "A_USER_ID",
"emails" : [
{
"address" : "AN_EMAIL_ADDRESS",
"verified" : false
}
],
"profile" : {
"name" : "NAME",
"company" : "A COMPANY",
"title" : "A TITLE",
"phone" : "123-123-1234",
"disabled" : false,
"universities" : [
{
"_id" : "ID_1",
"name" : "UNIVERSITY_NAME_1",
"code" : "CODE_1",
"styles" : {AN_OBJECT}
},
{
"_id" : "ID_2",
"name" : "UNIVERSITY_NAME_2",
"code" : "CODE_2",
"styles" : {AN_OBJECT}
}
]
}
}
是否可以使用其他运算符代替 $group
来实现此目的?还是我理解错了 $group
的目的?
编辑:这是原文post,上下文:
因为 $lookup
operator produces an array field, you need to $unwind
新字段在 $group
管道之前得到想要的结果:
db.users.aggregate([
{ "$unwind": "$profile" },
{ "$unwind": {
"path": "$profile.universities",
"preserveNullAndEmptyArrays": true
} },
{ "$lookup": {
"from": "universities",
"localField": "profile.universities._id",
"foreignField": "_id",
"as": "universities"
} },
{ "$unwind": "$universities" },
{ "$group": {
"_id": "$_id",
"emails": { "$first": "$emails" },
"profile": { "$first": "$profile" },
"universities": { "$push": "$universities" }
} },
{ "$project": {
"emails": 1,
"profile.name" : 1,
"profile.company": 1,
"profile.title" : 1,
"profile.phone" : 1,
"profile.disabled": 1,
"profile.universities": "$universities"
} }
]).pretty()
MongoDB 聚合在一分钟内变得呈指数级复杂化!
我正在 $unwind
一个嵌套数组,然后对展开的嵌套数组中的每个对象的 _id
执行 $lookup
。我最后的尝试是用 $group
反转平仓。但是,我无法重建原始的嵌入式数组,其原始 属性 名称以及每个文档的其余原始直接属性。
这是我目前的尝试:
db.users.aggregate([
{
$unwind: "$profile",
$unwind: {
path: "$profile.universities",
preserveNullAndEmptyArrays: true
}
},
{
$lookup: {
from: "universities",
localField: "profile.universities._id",
foreignField: "_id",
as: "profile.universities"
}
},
{
$group: {
_id: "$_id",
emails: { "$first": "$emails" },
profile: { "$first": "$profile" },
universities: { "$push": "$profile.universities" }
}
}
]).pretty()
我得到的是这样的:
{
"_id" : "A_USER_ID",
"emails" : [
{
"address" : "AN_EMAIL_ADDRESS",
"verified" : false
}
],
"profile" : {
"name" : "NAME",
"company" : "A COMPANY",
"title" : "A TITLE",
"phone" : "123-123-1234",
"disabled" : false,
"universities" : [
{
"_id" : "ID_1",
"name" : "UNIVERSITY_NAME_1",
"code" : "CODE_1",
"styles" : {AN_OBJECT}
}
]
},
"universities" : [
[
{
"_id" : "ID_1",
"name" : "UNIVERSITY_NAME_1",
"code" : "CODE_1",
"styles" : {AN_OBJECT}
}
],
[
{
"_id" : "ID_2",
"name" : "UNIVERSITY_NAME_2",
"code" : "CODE_2",
"styles" : {AN_OBJECT}
}
]
]
}
此结果有 2 个问题:
- 结果
universities
是一个数组数组,每个数组包含一个对象,因为$lookup
为原始$profile.universities
嵌套数组返回了单个元素数组。它应该只是一个对象数组。 - 生成的
universities
应该位于其嵌套在profiles
下的原始位置。我知道为什么原来的profile.universities
是这样的,因为我使用的是$first
运算符。我这样做的目的是保留profile
的所有原始属性,同时保留原始嵌套的universities
数组。
最终,我需要的是这样的:
{
"_id" : "A_USER_ID",
"emails" : [
{
"address" : "AN_EMAIL_ADDRESS",
"verified" : false
}
],
"profile" : {
"name" : "NAME",
"company" : "A COMPANY",
"title" : "A TITLE",
"phone" : "123-123-1234",
"disabled" : false,
"universities" : [
{
"_id" : "ID_1",
"name" : "UNIVERSITY_NAME_1",
"code" : "CODE_1",
"styles" : {AN_OBJECT}
},
{
"_id" : "ID_2",
"name" : "UNIVERSITY_NAME_2",
"code" : "CODE_2",
"styles" : {AN_OBJECT}
}
]
}
}
是否可以使用其他运算符代替 $group
来实现此目的?还是我理解错了 $group
的目的?
编辑:这是原文post,上下文:
因为 $lookup
operator produces an array field, you need to $unwind
新字段在 $group
管道之前得到想要的结果:
db.users.aggregate([
{ "$unwind": "$profile" },
{ "$unwind": {
"path": "$profile.universities",
"preserveNullAndEmptyArrays": true
} },
{ "$lookup": {
"from": "universities",
"localField": "profile.universities._id",
"foreignField": "_id",
"as": "universities"
} },
{ "$unwind": "$universities" },
{ "$group": {
"_id": "$_id",
"emails": { "$first": "$emails" },
"profile": { "$first": "$profile" },
"universities": { "$push": "$universities" }
} },
{ "$project": {
"emails": 1,
"profile.name" : 1,
"profile.company": 1,
"profile.title" : 1,
"profile.phone" : 1,
"profile.disabled": 1,
"profile.universities": "$universities"
} }
]).pretty()