laravel join：如何根据我的条件连接表并从同一个连接中提取各种数据？

Question

This is what I have done so far.

DB::table('products')
 ->leftjoin('inventories', 'inventories.product_id', 'products.id')
 })
 ->leftjoin('order_items as purchase_oi', function($query){
 $query->on('purchase_oi.inventory_id', 'inventories.id');
  $query->on('purchase_oi.status','!=', \DB::raw('"cancelled"'));
                        })                       
 ->select(DB::raw('sum(inventories.quantity)as qty'),
 DB::raw('count(purchase_oi.inventory_id) as purchases'),                            
  'products.id as pid', 'products.*')                          
  ->where('products.is_deleted', 0)
  ->where('inventories.is_deleted',0)
  ->groupBy('inventories.product_id')                  
  ->get();

order_items table contains the status in

enum('shipped','return''cancelled');

This issue is here on the join with order_items table. Currently, I'm just getting the data which is not equal to cancelled but I want to get the order_items data which are returned and the order_items data which are cancelled. i.e get me the purchases,cancelled and return count from order_items table. Expected output is:

0=>purchases = 4,
   returns = 10,
   cancelled = 2

Answer 1

I think the problem is your groupBy(). If you have the following query:

SELECT COUNT(*) AS count FROM table GROUP BY id

You don't end up with

+-------+
| count |
+-------+
| 5     |
+-------+

Instead you end up with:

+-------+
| count |
+-------+
| 1     |
+-------+
| 1     |
+-------+
| 1     |
+-------+
| 1     |
+-------+
| 1     |
+-------+

The count of all items per grouping.

Try this:

\DB::enableQueryLog();
\DB::flushQueryLog();

//Execute your query

$queries = /DB::getQueryLog();
/DB::disableQueryLog();
/DB::flushQueryLog();

$queries = collect($queries)->map(function (array $query) {
    return array_only($query, ['query', 'bindings']);
})->toArray();

dd($queries);

This should tell you exactly what SQL is being executed.