MongoDB - Get an unique array from two queries

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.