Mongoose populate and sort by length struggle - mongodb

Question

I have the following mongoose schema:
const userSchema = new mongoose.Schema({
email: { type: String, unique: true },
fragments: [{type: mongoose.Schema.Types.ObjectId, ref: 'Fragment'}]
}, { timestamps: true, collection: 'user' });
And
const fragmentSchema = new mongoose.Schema({
text: String,
owner: {type: mongoose.Schema.Types.ObjectId, ref: 'User'},
}, { timestamps: true, collection: 'fragment' });
In the data, I have a reference in the Fragment, but not in the User:
User:
{
"_id" : ObjectId("58373e571cbccb010012bfcd"),
"email" : "email#example.com",
// no "fragments": [ObjectId('58075ce37b7f2f01002b718f')] etc.
}
Fragment:
{
"_id" : ObjectId("58075ce37b7f2f01002b718f"),
"text" : "Donc, il faut changer de méthode",
"owner" : ObjectId("58075ce27b7f2f01002b717f")
}
I would like to query users sorted by the count of number of fragments, and I can't achieve this ...
First, I'd like to make this work:
User.find({_id: '58075ce27b7f2f01002b717f'})
.populate('fragments').exec(console.log)
returns
{
_id: 58075ce27b7f2f01002b717f,
email: 'bububg#hotmail.fr',
fragments: []
}
while I should have at least the above fragment included.
And regarding the sorted query, here's where I am now:
User.aggregate([
{ "$project": {
"email": 1,
"fragments": 1,
"nbFragments": { "$size": { "$ifNull": [ "$fragments", [] ] } }
}},
{ "$sort": { "nbFragments": -1 } }
], console.log)
At least it runs, but all the nbFragments fields are set to 0. This might be related to the fact that .populate('fragments') doesn't work but I can't be sure.
Thanks for the help, I did not expect so much trouble using Mongodb...
EDIT: thanks #Veeram, unfortunately your solution is not working:
User.find({}).find({_id: '58075ce27b7f2f01002b717f'}).populate('fragments').exec(console.log)
[ { _id: 58075ce27b7f2f01002b717f,
email: 'email#example.com',
// no fragments entry
} ]
while I updated my schema:
userSchema.virtual('fragments', {
ref: 'Fragment',
localField: '_id',
foreignField: 'owner',
options: { sort: { number: 1 }}, // Added sort just as an example
});
And regarding the aggregate, with:
User.aggregate([{
$lookup: {
from: 'Fragment',
localField: '_id',
foreignField: 'owner',
as: 'fragments'
}
}, { "$project": {
"email": 1,
"fragments": 1,
"nbFragments": {
"$size": { "$ifNull": [ "$fragments", [] ] } }
}}, { "$sort": { "nbFragments": -1 } }
]).exec(console.log)
I get:
{
_id: 58075ce27b7f2f01002b717f,
email: 'email#example.com',
fragments: [] // fragments are always empty while they shouldn't!
}

Answer 1

Tested with following data
User:
{
"_id" : ObjectId("58373e571cbccb010012bfcd"),
"email" : "email#example.com"
}
Fragment:
{
"_id" : ObjectId("58075ce37b7f2f01002b718f"),
"text" : "Donc, il faut changer de méthode",
"owner" : ObjectId("58373e571cbccb010012bfcd")
}
Response
[{"_id":"58373e571cbccb010012bfcd","email":"email#example.com","fragments":[{"_id":"58075ce37b7f2f01002b718f","text":"Donc, il faut changer de méthode","owner":"58373e571cbccb010012bfcd"}],"nbFragments":1}]
You define schema to use owner to populate the fragments also called virtual population. http://mongoosejs.com/docs/populate.html
const userSchema = new mongoose.Schema({
email: { type: String, unique: true }
}, { timestamps: true, collection: 'user' });
var User = mongoose.model("User", userSchema);
const fragmentSchema = new mongoose.Schema({
text: String,
owner: {type: mongoose.Schema.Types.ObjectId, ref: 'User'},
}, { timestamps: true, collection: 'fragment' });
var Fragment = mongoose.model("Fragment", fragmentSchema);
userSchema.virtual('fragments', {
ref: 'Fragment',
localField: '_id',
foreignField: 'owner',
options: { sort: { text: -1 }}, // Added sort just as an example
});
This will now work as expected, but I don't know a way to sort on some dynamic field like count of number of fragments in mongoose. I don't think it is possible
User.find({_id: '58373e571cbccb010012bfcd'})
.populate('fragments').exec(function (err, user) {
console.log(JSON.stringify(user));
});
Okay now for dynamic sorting, you have to use alternative raw mongo query with a $lookup (equivalent of populate).
const userSchema = new mongoose.Schema({
email: { type: String, unique: true }
}, { timestamps: true, collection: 'user' });
var User = mongoose.model("User", userSchema);
const fragmentSchema = new mongoose.Schema({
text: String,
owner: {type: mongoose.Schema.Types.ObjectId, ref: 'User'},
}, { timestamps: true, collection: 'fragment' });
User.aggregate([{
$lookup: {
from: 'fragment',
localField: '_id',
foreignField: 'owner',
as: 'fragments'
}
}, { "$project": {
"email": 1,
"fragments": 1,
"nbFragments": {
"$size": { "$ifNull": [ "$fragments", [] ] } }
}}, { "$sort": { "nbFragments": -1 } }
]).exec(function (err, user) {
console.log(JSON.stringify(user));
})