MongoDB , getting the minimum & maximum of array subset - mongodb

Question

Am trying to find a way to get the minimum number of orders between
2019-03-17 and 2019-03-19 excluding 2019-03-15 from the results ..
{
"_id" : ObjectId("5c8ffdadde62bf097d54ec47"),
"productId" : "32886845998",
"orders" : [
{
"date" : ISODate("2019-03-15T00:00:00.000+0000"),
"orders" : NumberInt(9)
},
{
"date" : ISODate("2019-03-17T00:00:00.000+0000"),
"orders" : NumberInt(21)
},
{
"date" : ISODate("2019-03-18T00:00:00.000+0000"),
"orders" : NumberInt(20)
},
{
"date" : ISODate("2019-03-19T00:00:00.000+0000"),
"orders" : NumberInt(30)
}
]
}
I tried using $min and $max operator but that didn't help because it iterated through the full array to find maximum & minimum
db.products.aggregate([
{
$project: {
maximum: {
$reduce: {
input: "$orders",
initialValue: 0,
in: {
$max: [
"$$value",
{
$cond: [
{ $gte: [ "$$this.date", ISODate("2019-03-17T00:00:00.000+0000") ] },
"$$this.orders",
0
]
}
]
}
}
}
}
}
])

Answer 1

You can use $filter to apply filtering by orders.date and then you can apply $min and $max on filtered set:
db.col.aggregate([
{
$project: {
filteredOrders: {
$filter: {
input: "$orders",
cond: {
$and: [
{ $gte: [ "$$this.date", ISODate("2019-03-17T00:00:00.000+0000") ] },
{ $lte: [ "$$this.date", ISODate("2019-03-19T00:00:00.000+0000") ] },
]
}
}
}
}
},
{
$project: {
min: { $min: "$filteredOrders.orders" },
max: { $max: "$filteredOrders.orders" },
}
}
])