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?我对如何在这种类型的查询中使用Laravel连接查询构建器感到困惑。
答案!!
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();
答案 0 :(得分:2)
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();
答案 1 :(得分:1)
使用查询构建器很难构建这种类型的查询。但是,您可以使用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");
如果需要将参数与查询绑定:
$deleted = 0;
DB::select("SELECT a.id, active, name, email, img_location, IFNULL(b.Total, 0) AS LeadTotal, IFNULL(c.Total, 0) AS InventoryTotal
FROM users AS a
LEFT JOIN (
SELECT user_id, count(*) as Total
FROM lead_user
GROUP BY user_id
) AS b ON a.id = b.user_id
LEFT JOIN (
SELECT user_id, count(*) as Total
FROM user_inventory
GROUP BY user_id
) AS c ON a.id = c.user_id
WHERE a.is_deleted = ?", [$deleted]);
答案 2 :(得分:0)
我相信,使用ORM关系是连接表的更好方法。