我的表格如下:
商店
[id]
库存
[id, shop_id]
订单
[id, shop_id]
order_item
[order_id, inventory_id, quantity]
型号:
//Shop
class Shop extends Model
{
public function inventories()
{
return $this->hasMany(Inventory::class);
}
public function orders()
{
return $this->hasMany(Order::class);
}
}
//Inventory
class Inventory extends Model
{
public function shop()
{
return $this->belongsTo(Shop::class);
}
public function orders()
{
return $this->belongsToMany(Order::class, 'order_items')
->withPivot('quantity');
}
}
//Order
class Order extends Model
{
public function shop()
{
return $this->belongsTo(Shop::class);
}
public function inventories()
{
return $this->belongsToMany(Inventory::class, 'order_items')
->withPivot('quantity');
}
}
现在我想要一个特定商店的5个畅销库存,最好的方法是什么?
我在使用Laravel 5.5
答案 0 :(得分:1)
select s.id,sum(oi.quantity) as total from munna.shops as s
join munna.inventories as iv on s.id=iv.shop_id
join munna.orders as o on iv.shop_id=o.shop_id
join munna.order_items as oi on o.id=oi.order_id
group by s.id
order by total desc limit 5
答案 1 :(得分:1)
首先,通过查看 order_item 上的表格, ,order_id和inventory_id 肯定会来到同一家商店吗?我想是的,因为如果没有,你会有2个不同的商店,顶级订单相同。我不知道你为什么这样做,但它有点令人困惑,不能弄明白为什么,但我会尝试这个:
public function topOrders()
{
$items = DB::table('shops')
->join('orders', 'shops.id', '=', 'orders.shop_id')
->join('inventories', 'shops.id', '=', 'inventories.shop_id')
->join('order_items', 'orders.id', '=', 'order_items.order_id')
->orderBy('quantity', 'desc')
->take(5)
->get();
return $items;
}
我写的内容应该从所有3行中选择所有内容,如果你只想选择你想要选择的项目或任何你可以指定它添加一个select子句
答案 2 :(得分:1)
虽然这是我自己的问题,但我自己找到了解决方案,并希望与社区分享解决方案。我想用Eloquent解决它,因为我需要视图上的模型,并且不想再次查询模型。
Inventory::where('shop_id', \Auth::user()->shop_id)
->select(
'inventories.*',
\DB::raw('SUM(order_items.quantity) as quantity')
)
->join('order_items', 'inventories.id', 'order_items.inventory_id')
->groupBy('inventory_id')
->get();
我希望这会帮助有类似问题的人。感谢