Laravel 4查询构建器 - 具有复杂的左连接

Question

我是Laravel 4的新手。

我有这个问题：

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

如何将其转换为Laravel查询构建器？我对如何将Laravel连接查询构建器与此类查询一起使用感到困惑。

回答！！

请问lakvel论坛上所有petkostas的帮助。我们得到了答案。

$users = DB::table('users AS a')
->select(array('a.*', DB::raw('IFNULL(b.Total, 0) AS LeadTotal'), DB::raw('IFNULL(c.Total, 0) AS InventoryTotal')  ) )
->leftJoin(DB::raw('(SELECT user_id, COUNT(*) as Total FROM lead_user GROUP BY user_id) AS b'), function( $query ){
    $query->on( 'a.id', '=', 'b.user_id' );
})
->leftJoin(DB::raw('(SELECT user_id, COUNT(*) as Total FROM user_inventory WHERE is_deleted = 0 GROUP BY user_id) AS c'), function( $query ){
    $query->on( 'a.id', '=', 'c.user_id' );
})
->where('a.is_deleted', '=', 0)
->get();

Answer 1

我相信这应该有效：

$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();

Answer 2

使用查询构建器很难构建此类查询。但是，你可以使用DB::select

如果您无需绑定，可以使用以下内容：

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");

如果需要将参数绑定到查询：

$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]);

Answer 3

我相信，使用ORM关系是加入表格的更好方法。

请参考链接：https://laravel.com/docs/5.7/eloquent-relationships