How to compare two fields of one subdocument in array? - mongodb

Question

Given some documents in MongoDB
[
{
"id": 1,
"items": [
{
"id": "T0001",
"x": 10,
"y": 10
},
{
"id": "T0002",
"x": 10,
"y": 5
},
{
"id": "T0003",
"x": 10,
"y": 10
},
{
"id": "T0004",
"x": 10,
"y": 20
}
]
},
{
"id": 2,
"items": [
{
"id": "T0001",
"x": 10,
"y": 5
},
{
"id": "T0002",
"x": 10,
"y": 15
}
]
}
]
I would like to remove the subdocuments with items.id="T0001" and items.id="T0002" in document id=1. This could be done with the following command.
db.collection.update({
id: 1
},
{
$pull: {
items: {
id: {
$in: [
"T0001",
"T0002"
]
}
}
}
})
However, if I would like to add one more condition "items.x === items.y" into the $pull operation. (i.e. only the subdocument whose id="T0001" will be removed because its x=10 and y=10)
The sample code can be found here https://mongoplayground.net/p/rYJp8wQPpcS.
Could anyone show me some tips to solve this problem? Thank you!

Answer 1

Working on the update with the aggregation pipeline will be easier.
$set - Set the items field.
1.1. $filter - Filter the items array by condition:
1.1.1. $not - Negate the result of 1.1.1.1.
1.1.1.1 $and - Fulfill both conditions of id in ["T0001", "T0002"] and x and y are equal.
db.collection.update({
id: 1
},
[
{
$set: {
items: {
$filter: {
input: "$items",
cond: {
$not: {
$and: [
{
$in: [
"$$this.id",
[
"T0001",
"T0002"
]
]
},
{
$eq: [
"$$this.x",
"$$this.y"
]
}
]
}
}
}
}
}
}
])
Sample Mongo Playground