英文:
Double array $lookup call preserving original structure
问题
我尝试在聚合中填充一个双重数组对象，所以我正在使用`$lookup`函数。集合看起来像这样：
{
foo: [
{
bar: [
{
_id: ObjectId('63f508eedd2962118c37ea36')
}
]
}
]
}
我的lookup如下：
{
$lookup: {
from: "collection",
localField: "foo.bar",
foreignField: "_id",
as: "foo.bar",
},
}
这导致了以下结果：
{
foo: {
bar: [
{
_id: ObjectId('63f508eedd2962118c37ea36'),
field1: "helloworld"
}
]
}
}
而我实际想要的是：
{
foo: [
{
bar: [
{
_id: ObjectId('63f508eedd2962118c37ea36'),
field1: "helloworld"
}
]
}
]
}
有没有关于如何在聚合中实现我想要的效果的想法？
<details>
<summary>英文:</summary>
I'm trying to populate a double array object but in an aggregate so I am utilizing the `$lookup` function. The collection looks something like this:
```BSON
{
foo: [
{
bar: [
{
_id: ObjectId('63f508eedd2962118c37ea36')
}
]
}
]
}
My lookup looks like:
{
$lookup: {
from: "collection",
localField: "foo.bar",
foreignField: "_id",
as: "foo.bar",
},
}
which results in
{
foo: {
bar: [
{
_id: ObjectId('63f508eedd2962118c37ea36'),
field1: "helloworld"
}
]
}
}
where what I actually want is
{
foo: [
{
bar: [
{
_id: ObjectId('63f508eedd2962118c37ea36'),
field1: "helloworld"
}
]
}
]
}
Any ideas on how to achieve what I want in an aggregate?
答案1
得分: 1
-
$lookup- 使用$lookup来连接collection集合并返回bars数组字段。 -
$set- 设置foo数组。2.1.
$map- 遍历foo数组中的元素。2.1.1.
$mergeObjects- 将当前迭代的foo元素与2.1.1.1.1的结果合并。2.1.1.1. 包含结果为2.1.1.1.1.1的
bar字段的文档。2.1.1.1.1.
$map- 遍历bar数组中的元素。2.1.1.1.1.1.
$mergeObjects- 将当前迭代的bar元素与2.1.1.1.1.1.1的结果合并。2.1.1.1.1.1.1.
$first- 从bars数组中按_id匹配获取第一个元素。 -
$unset- 移除bars数组。
英文:
Seems a direct map with $lookup for the nested array is not possible.
-
$lookup- Join thecollectioncollection and return thebarsarray field. -
$set- Setfooarray.2.1.
$map- Iterate element infooarray.2.1.1.
$mergeObjects- Merge current iteratedfooelement with the result of 2.1.1.1.2.1.1.1. A document with a
barfield which contains the result of 2.1.1.1.1.2.1.1.1.1.
$map- Iterate element inbararray.2.1.1.1.1.1.
$mergeObjects- Merge the current iteratedbarelement with the result of 2.1.1.1.1.1.1.2.1.1.1.1.1.1.
$first- Get the first matching element from thebarsarray by_ids. -
$unset- Removebarsarray.
db.from.aggregate([
{
$lookup: {
from: "collection",
localField: "foo.bar._id",
foreignField: "_id",
as: "bars"
}
},
{
$set: {
foo: {
$map: {
input: "$foo",
as: "foo",
in: {
$mergeObjects: [
"$$foo",
{
bar: {
$map: {
input: "$$foo.bar",
as: "bar",
in: {
$mergeObjects: [
"$$bar",
{
$first: {
$filter: {
input: "$bars",
cond: {
$eq: [
"$$bar._id",
"$$this._id"
]
}
}
}
}
]
}
}
}
}
]
}
}
}
}
},
{
$unset: "bars"
}
])
答案2
得分: 1
也许在 foo 层级上首先使用 $unwind，然后执行 $lookup 和 $group 会更简单？
db.from.aggregate([
{
"$unwind": "$foo"
},
{
$lookup: {
from: "collection",
localField: "foo.bar._id",
foreignField: "_id",
as: "foo.bar"
}
},
{
$group: {
_id: "$_id",
foo: {
$push: "$foo"
}
}
}
])
英文:
Maybe it would be simpler to $unwind at foo level first, do the $lookup and $group back the result?
db.from.aggregate([
{
"$unwind": "$foo"
},
{
$lookup: {
from: "collection",
localField: "foo.bar._id",
foreignField: "_id",
as: "foo.bar"
}
},
{
$group: {
_id: "$_id",
foo: {
$push: "$foo"
}
}
}
])
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