Aggregate function two match with lookup mongodb - mongodb

Question

db.setting.aggregate([
{
$match: {
status: true,
deleted_at: 0,
_id: {
$in: [
ObjectId("5c4ee7eea4affa32face874b"),
ObjectId("5ebf891245aa27c290672325")
]
}
}
},
{
$lookup: {
from: "site",
localField: "_id",
foreignField: "admin_id",
as: "data"
}
},
{
$project: {
name: 1,
status: 1,
numberOfRecord: {
$size: "$data"
}
}
},
{
$sort: {
numberOfRecord: 1
}
}
])
I would like to fetch record and number of record which are greater then equal to 2020-01-01 tried to add below code but doesn't get success.
How can I make this correct please guide thanks in advance. Here is playground https://mongoplayground.net/p/GU8WbTVqo2I
{
$match: {
"data.createdAt": {
$gte: new Date("2020-01-01")
}
}
},
Output should be
[
{
"_id": ObjectId("5ebf891245aa27c290672325"),
"name": "Menz",
"numberOfRecord": 0,
"status": true
},
{
"_id": ObjectId("5c4ee7eea4affa32face874b"),
"name": "Dave",
"numberOfRecord": 1, // instead 2 bcoz this is only gte "2020-01-01"
"status": true
}
]

Answer 1

You can use $unwind to split up the array, but using $match will completely eliminate documents that don't have any matching documents, so you would need to used $group with a conditional count, perhaps:
db.setting.aggregate([
{$match: {
status: true,
deleted_at: 0,
_id: {
$in: [
ObjectId("5c4ee7eea4affa32face874b"),
ObjectId("5ebf891245aa27c290672325")
]
}
}},
{$lookup: {
from: "site",
localField: "_id",
foreignField: "admin_id",
as: "data"
}},
{$unwind: {
path: "$data",
preserveNullAndEmptyArrays: true
}},
{$group: {
_id: "$_id",
name: {$first: "$name"},
status: {$first: "$status"},
numberOfRecord: {
$sum:{
$cond:{
if:{
$gte:[
"$data.createdAt",
new Date("2020-01-01")
]
},
then: 1,
else: 0
}
}
}
}},
{$sort: { numberOfRecord: 1 }}
])
Playground