MongoDB - Get an unique array from two queries - database

Question

First of all sorry if the title does not reflect what I really need to do.
Imagine a collection that represents what products sell which vendor. Let´s simplify like:
{_id, productID, VendorID, Price, Stock}
For example, we could have:
{_id: 1 ,productID: 1, vendorID: A, price: 0, stock: 0},
{_id: 2 ,productID: 2, vendorID: A, price: 0, stock: 0},
{_id: 3 ,productID: 3, vendorID: A, price: 0, stock: 0},
{_id: 4 ,productID: 4, vendorID: A, price: 0, stock: 0},
{_id: 5 ,productID: 1, vendorID: B, price: 0, stock: 19},
{_id: 6 ,productID: 3, vendorID: B, price: 0, stock: 21}
The idea is that Vendor A is the Super Admin and the one who controls which products can be sold in the marketplace. That´s why all products sold by A have price = 0 and stock = 0.
What I am trying to get within the same query is. When Vendor B is logged:
List all the items that A is selling.
But if a product is sold by B, then I should return it instead of the A-one.
The result would look like:
{_id: 2 ,productID: 2, vendorID: A, price: 0, stock: 0},
{_id: 4 ,productID: 4, vendorID: A, price: 0, stock: 0},
{_id: 5 ,productID: 1, vendorID: B, price: 0, stock: 19},
{_id: 6 ,productID: 3, vendorID: B, price: 0, stock: 21}
Do you have any idea how I can do it in a shot?
I can do the filter in the frontend but I would prefer doing like this in order to avoid problems with pagination for example.

Answer 1

If you are sure that all products sold by A always have price = 0 and stock = 0, you can do the following trick:
db.collection.aggregate({
$match: {
vendorID: {
$in: [
"A",
"B"
]
}
}
},
{
$sort: {
productID: 1,
price: -1
}
},
{
"$group": {
"_id": "$productID",
"product": {
$first: "$$ROOT"
}
}
})
Mongo playground
by $match you are looking for all products owned by default vendor (A) and target vendor (B).
by $sort you group products by productID and by price in each group, so product sold by A will be last one in each group
by $group you extract product from each group
using $first it will be first product (with higher price, so product sold by A will be matched only if no other products with same ID were sold).
DB documents will be aggregated by $$ROOT into product field, so you can map them using:
// (javascript)
const productDocuments = aggregationResult.map(row => row.product);
Notice: product sold by A still can be extracted even if B sold product with same ID in cases when B's product also have price = 0.