Display best selling items by status - php

Question

Explaining my problem, I have two tables in my database called order and order_details.
On my dashboard, I display the three best-selling items (currently work!). However, I would like to display only the best selling items THAT have the status = delivered.
Today, it works like this:
$top_sell_items = OrderDetails::with(['product'])
->select('product_id', DB::raw('SUM(quantity) as count'))
->groupBy('product_id')
->orderBy("count", 'desc')
->take(3)
->get();
The problem is that the order status is stored in another table, called orders, column order_status.
How can I create this rule and include it in my $top_sell_items?

Answer 1

if you relationship is done between this table already, you can use this code, if not you have to go to the OrderDetails Model and add new method orders
$top_sell_items = OrderDetails::with(['product', 'orders'])
->whereHas('orders', function($query) {
$query->where('status', 'delivered');
})
->select('product_id', DB::raw('SUM(quantity) as count'))
->groupBy('product_id')
->orderBy('count', 'desc')
->take(3)
->get();

Answer 2

You could either define a relationship between Orders and OrderDetails or use a join like so...
<?php
$top_sell_items = OrderDetails::with(['product'])
->join('orders', 'orders.id', '=', 'order_details.order_id')
->select('product_id', DB::raw('SUM(quantity) as count'))
->where('orders.order_status', 'delivered');
->groupBy('product_id')
->orderBy("count", 'desc')
->take(3)
->get();
More info here: https://laravel.com/docs/8.x/queries#joins

Answer 3

Depending if you desire this, the following solution might be the most efficient:
$products = Product
::whereHas('orderDetails.order', function ($query) {
$query->where('orders.order_status', 'delivered');
})
->withSum('orderDetails', 'quantity')
->orderBy('order_details_sum_quantity', 'desc')
->take(3)
->get();
It will directly return instances of Product. In addition it puts everything in a single query instead of the two that with produces.