问题描述
我正在与MongoDB中新的(可爱的)lookup
运算符作斗争.我有3个收藏集:
I am struggling with the newish (lovely) lookup
operator in MongoDB. I have 3 collections:
艺术家
{
"_id" : ObjectId("5b0d2b2c7ac4792df69a9942"),
"name" : "Dream Theater",
"started_in" : NumberInt(1985),
"active" : true,
"country" : "US",
"current_members" : [
ObjectId("5b0d2a7c7ac4792df69a9941")
],
"previous_members" : [
ObjectId("5b0d2bf57ac4792df69a9954")
],
"albums" : [
ObjectId("5b0d16ee7ac4792df69a9924"),
ObjectId("5b0d47667ac4792df69a9994")
],
"genres" : [
"prog metal",
"prog rock"
]
}
相册
{
"_id" : ObjectId("5b0d16ee7ac4792df69a9924"),
"title" : "Images and words",
"released" : ISODate("1992-07-07T00:00:00.000+0000"),
"songs" : [
ObjectId("5b0d15ab7ac4792df69a9916"),
ObjectId("5b0d15ee7ac4792df69a991e"),
ObjectId("5b0d2db37ac4792df69a995d"),
ObjectId("5b0d2dbe7ac4792df69a995e"),
ObjectId("5b0d2dcb7ac4792df69a995f"),
ObjectId("5b0d2dd87ac4792df69a9960"),
ObjectId("5b0d2de27ac4792df69a9961"),
ObjectId("5b0d2dec7ac4792df69a9962")
],
"type" : "LP"
}
{
"title" : "Awake",
"released" : ISODate("1994-10-04T00:00:00.000+0000"),
"songs" : [
ObjectId("5b0d470d7ac4792df69a9991")
],
"type" : "LP",
"_id" : ObjectId("5b0d47667ac4792df69a9994")
}
歌曲
{
"_id" : ObjectId("5b0d15ab7ac4792df69a9916"),
"title" : "Pull me under"
}
{
"_id" : ObjectId("5b0d15ee7ac4792df69a991e"),
"title" : "Another day"
}
{
"title" : "Take the time",
"_id" : ObjectId("5b0d2db37ac4792df69a995d")
}
{
"title" : "Surrounded",
"_id" : ObjectId("5b0d2dbe7ac4792df69a995e")
}
{
"title" : "Metropolis - part I",
"_id" : ObjectId("5b0d2dcb7ac4792df69a995f")
}
{
"title" : "Under a glass moon",
"_id" : ObjectId("5b0d2dd87ac4792df69a9960")
}
{
"title" : "Wait for sleep",
"_id" : ObjectId("5b0d2de27ac4792df69a9961")
}
{
"title" : "Learning to live",
"_id" : ObjectId("5b0d2dec7ac4792df69a9962")
}
{
"title" : "6:00",
"_id" : ObjectId("5b0d470d7ac4792df69a9991")
}
我可以轻松地使用$lookup
进行聚合以获取详细的albums
数组,但是如何在相应的相册中获取详细的songs
呢?我想扩展以下查询:
I can easily do an aggregation with $lookup
to get the detailed albums
array, but how do I get also the detailed songs
in the corresponding albums?I would like to extend the following query:
db.artists.aggregate([ {
$lookup: {
from: "albums",
localField: "albums",
foreignField: "_id",
as: "albums"
}
}]).pretty()
推荐答案
如果您的mongodb版本为 3.6 ,则可以尝试使用嵌套的 $lookup
聚合...
If you have mongodb version 3.6 then you can try with nested $lookup
aggregation...
db.collection.aggregate([
{ "$lookup": {
"from": Albums.collection.name,
"let": { "albums": "$albums" },
"pipeline": [
{ "$match": { "$expr": { "$in": [ "$_id", "$$albums" ] } } },
{ "$lookup": {
"from": Songs.collection.name,
"let": { "songs": "$songs" },
"pipeline": [
{ "$match": { "$expr": { "$in": [ "$_id", "$$songs" ] } } }
],
"as": "songs"
}}
],
"as": "albums"
}}
])
对于冗长的解释,您可以通过 $ lookup多个级别而无需$ unwind?
And for long-winded explanation you can go through $lookup multiple levels without $unwind?
或者如果您的Mongodb版本低于 3.6
Or If you have mongodb version prior to 3.6
db.collection.aggregate([
{ "$lookup": {
"from": Albums.collection.name,
"localField": "albums",
"foreignField": "_id",
"as": "albums"
}},
{ "$unwind": "$albums" },
{ "$lookup": {
"from": Songs.collection.name,
"localField": "albums.songs",
"foreignField": "_id",
"as": "albums.songs",
}},
{ "$group": {
"_id": "$_id",
"name": { "$first": "$name" },
"started_in": { "$first": "$started_in" },
"active": { "$first": "$active" },
"country": { "$first": "$country" },
"albums": {
"$push": {
"_id": "$albums._id",
"title": "$albums.title",
"released": "$albums.released",
"type": "$albums.type",
"songs": "$albums.songs"
}
}
}}
])
这篇关于$ lookup mongodb中的嵌套数组的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!