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Mysql Laravel 4:DB::在状态中选择_Mysql_Laravel_Laravel 4 - Fatal编程技术网

Mysql Laravel 4:DB::在状态中选择

Mysql Laravel 4:DB::在状态中选择,mysql,laravel,laravel-4,Mysql,Laravel,Laravel 4,运行laravel DB::在Station中选择时,如文档所示: //$list_of_ids = '1,3,5,66,3' //$uid = 1 DB::select(' SELECT SUM(score) FROM user WHERE id IN (?) and user_id = ? LIMIT 4 ', array($list_of_ids,$uid)); 没有收到任何结果 但在运行时: DB::select(' SELEC

运行laravel DB::在Station中选择时,如文档所示:

//$list_of_ids = '1,3,5,66,3'
//$uid = 1

DB::select('
    SELECT SUM(score)
    FROM user
    WHERE id
    IN (?)
    and user_id = ?
    LIMIT 4
', array($list_of_ids,$uid));
没有收到任何结果

但在运行时:

DB::select('
    SELECT SUM(score)
    FROM user
    WHERE id
    IN ('.$list_of_ids.')  //add var here
    and user_id = ?
    LIMIT 4
', array($uid));    

返回正确的结果,第一种方法是什么?

您需要的是使用实际数组和正确的查询

$userIds = explode(',', $list_of_ids);
DB::table('user')->whereIn('id', $userIds)->where('user_id', $uid)->limit(4)->get([DB::raw('SUM(score)')]);

这里有一个解释为什么这不起作用。简短的版本是,它与Laravel无关,而是PDO本身。