问题
so I have a collection users
, and each document in this collection, as well as other properties, has an array of ids of documents in the other collection: workouts
.
Every document in the collection workouts
has a property named date
.
And here's what I want to get:
For a specific user, I want to get an array of {workoutId, workoutDate} for the workouts that belong to that user, sorted by date.
This is my attempt, which is working fine.
Users.aggregate([
{
$match : {
_id : ObjectId("whateverTheUserIdIs")
}
},
{
$unwind : {
path : "$workouts"
}
}, {
$lookup : {
from : "workouts",
localField : "workouts",
foreignField : "_id",
as : "workoutDocumentsArray"
}
}, {
$project : {
_id : false,
workoutData : {
$arrayElemAt : [
$workoutDocumentsArray,
0
]
}
}
}, {
$project : {
date : "$workoutData.date",
id : "$workoutData._id"
}
}, {
$sort : {date : -1}
}
])
However I refuse to believe I need all this for what would be such a simple query in SQL!? I believe I must at least be able to merge the two $project
stages into one? But I've not been able to figure out how looking at the docs.
Thanks in advance for taking the time! ;)
==== EDIT - This is some sample data
Collection users:
[{
_id:xxx,
workouts: [2,4,6]
},{
_id: yyy,
workouts: [1,3,5]
}]
Colleciton workouts:
[{
_id:1,
date: 1/1/1901
},{
_id:2,
date: 2/2/1902
},{
_id:3,
date: 3/3/1903
},{
_id:4,
date: 4/4/1904
},{
_id:5,
date: 5/5/1905
},{
_id:6,
date: 6/6/1906
}]
And after running my query, for example for user xxx, I would like to get only the workouts that belong to him (whose ids appear in his workouts array), so the result I want would look like:
[{
id:6,
date: 6/6/1906
},{
id:4,
date: 4/4/1904
},{
id:2,
date: 2/2/1902
}]
回答1:
You don't need to $unwind the workouts
array as it already contains array of _id
s and use $replaceRoot instead of doing $project
Users.aggregate([
{ "$match": { "_id" : ObjectId("whateverTheUserIdIs") }},
{ "$lookup": {
"from" : "workouts",
"localField" : "workouts",
"foreignField" : "_id",
"as" : "workoutDocumentsArray"
}},
{ "$unwind": "$workoutDocumentsArray" },
{ "$replaceRoot": { "newRoot": "$workoutDocumentsArray" }}
{ "$sort" : { "date" : -1 }}
])
or even with new $lookup syntax
Users.aggregate([
{ "$match" : { "_id": ObjectId("whateverTheUserIdIs") }},
{ "$lookup" : {
"from" : "workouts",
"let": { "workouts": "$workouts" },
"pipeline": [
{ "$match": { "$expr": { "$in": ["$_id", "$$workouts"] }}},
{ "$sort" : { "date" : -1 }}
]
"as" : "workoutDocumentsArray"
}},
{ "$unwind": "$workoutDocumentsArray" },
{ "$replaceRoot": { "newRoot": "$workoutDocumentsArray" }}
])
来源:https://stackoverflow.com/questions/52720716/lookup-and-sort-the-foreign-collection