Laravel 4 query builder - with complicated left joins

我的梦境 提交于 2019-12-04 03:29:58

I believe this should work:

$users = DB::table('users')
    ->select( array('users.*', DB::raw('COUNT(lead_user.user_id) as LeadTotal'), DB::raw('COUNT(user_inventory.user_id) as InventoryTotal') ) )
    ->leftJoin('lead_user', 'users.id', '=', 'lead_user.user_id')
    ->leftJoin('user_inventory', 'users.id', '=', 'user_inventory.user_id')
    ->where('users.is_deleted', '=', 0)
    ->get();

This type of query is very hard to build with query builder. However, you can use DB::select

if you have nothing to bind, you can use following:

DB::select("SELECT a.id, active, name, email, img_location, IFNULL(b.Total, 0) AS LeadTotal, IFNULL(c.Total, 0) AS InventoryTotal
FROM users AS a
LEFT JOIN (
   SELECT user_id, count(*) as Total
   FROM lead_user
   GROUP BY user_id
) AS b ON a.id = b.user_id
LEFT JOIN (
   SELECT user_id, count(*) as Total
   FROM user_inventory
   GROUP BY user_id
) AS c ON a.id = c.user_id
WHERE a.is_deleted = 0");

if need to bind parameter with the query:

$deleted = 0;

DB::select("SELECT a.id, active, name, email, img_location, IFNULL(b.Total, 0) AS LeadTotal, IFNULL(c.Total, 0) AS InventoryTotal
FROM users AS a
LEFT JOIN (
   SELECT user_id, count(*) as Total
   FROM lead_user
   GROUP BY user_id
) AS b ON a.id = b.user_id
LEFT JOIN (
   SELECT user_id, count(*) as Total
   FROM user_inventory
   GROUP BY user_id
) AS c ON a.id = c.user_id
WHERE a.is_deleted = ?", [$deleted]);

I believe, use of ORM relations is a better way to join the tables.

please refer the link: https://laravel.com/docs/5.7/eloquent-relationships

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!