Reputation: 613
I need to fetch distinct nested documents.
Please find the sample document:
{
"propertyId": 1001820437,
"date": ISODate("2020-07-17T00:00:00.000Z"),
"HList":[
{
"productId": 123,
"name": "Dubai",
"tsh": true
}
],
"PList":[
{
"productId": 123,
"name": "Dubai",
"tsh": false
},
{
"productId": 234,
"name": "India",
"tsh": true
}
],
"CList":[
{
"productId": 234,
"name": "India",
"tsh": false
}
]
}
Expected result is:
{
"produts":[
{
"productId": 123,
"name": "Dubai"
},
{
"productId": 234,
"name": "India"
}
]
}
I tried with this query:
db.property.aggregate([
{
$match: {
"propertyId": 1001820437,
"date": ISODate("2020-07-17T00:00:00.000Z")
}
},
{
"$project": {
"_id": 0,
"unique": {
"$filter": {
"input": {
"$setDifference": [
{
"$concatArrays": [
"$HList.productId",
"$PList.productId",
"$CList.productId"
]
},
[]
]
},
"cond": {
"$ne": [ "$$this", "" ]
}
}
}
}
}
]);
Is $setDifference
aggregation is correct choice here?
My query returns only unique product ids but i need a productId
with name
.
Could someone help me to solve this?
Thanks in advance
Upvotes: 6
Views: 479
Reputation: 613
db.collection.aggregate([
{
$match: {
"propertyId": 1001820437,
"date": ISODate("2020-07-17T00:00:00.000Z")
}
},
{
$project: {
products: {
$filter: {
input: { "$setUnion" : ["$CList", "$HList", "$PList"] },
as: 'product',
cond: {}
}
}
}
},
{
$project: {
"_id":0,
"products.tsh": 1,
"products.name": 1,
}
},
])
Upvotes: 0
Reputation: 14287
The following two aggregations return the expected and same result (you can use any of the two):
db.collection.aggregate( [
{
$project: {
_id: 0,
products: {
$reduce: {
input: { $setUnion: [ "$HList", "$PList", "$CList" ] },
initialValue: [],
in: {
$setUnion: [ "$$value", [ { productId: "$$this.productId", name: "$$this.name" } ] ]
}
}
}
}
}
] )
This one is little verbose:
db.collection.aggregate( [
{
$project: { list: { $setUnion: [ "$HList", "$PList", "$CList" ] } }
},
{
$unwind: "$list"
},
{
$group: {
_id: null,
products: { $addToSet: { "productId": "$list.productId", "name": "$list.name" } }
}
},
{
$project: { _id: 0 }
}
] )
Upvotes: 4
Reputation: 49945
You can use $project
first to get rid of tsh
field and then run $setUnion which ignores duplicated entries:
db.collection.aggregate([
{
$project: {
"HList.tsh": 0,
"PList.tsh": 0,
"CList.tsh": 0,
}
},
{
$project: {
products: {
$setUnion: [ "$HList", "$PList", "$CList" ]
}
}
}
])
Upvotes: 5