Php Laravel 5执行自己的查询

Php Laravel 5执行自己的查询,php,laravel,Php,Laravel,是否有可能在Laravel 5中执行此查询 select t1.*, v.hash from ( select dr.*, d.id as d_id, d.name, d.user_id from directory d left join directory_relations dr on d.id = dr.directory_id union select dr.*, d.id as d_id, d.name, d.user_id from directory

是否有可能在Laravel 5中执行此查询

select t1.*, v.hash from 
    ( select dr.*, d.id as d_id, d.name, d.user_id from directory d 
    left join directory_relations dr on d.id = dr.directory_id 
    union select dr.*, d.id as d_id, d.name, d.user_id from directory_relations dr 
    left join directory d on d.id = dr.directory_id ) t1 
left join videos v on t1.video_id = v.id WHERE t1.user_id = 265
您可以使用DB::raw

$results = DB::select( DB::raw("select t1.*, v.hash from 
    ( select dr.*, d.id as d_id, d.name, d.user_id from directory d 
    left join directory_relations dr on d.id = dr.directory_id 
    union select dr.*, d.id as d_id, d.name, d.user_id from directory_relations dr 
    left join directory d on d.id = dr.directory_id ) t1 
left join videos v on t1.video_id = v.id WHERE t1.user_id = :user_id"), array(
   'user_id' => 265,
 ));

使用查询生成器,您可以执行以下操作:

$union = DB::select(['dr.*', 'd.id as d_id', 'd.name', 'd.user_id'])
    ->from('directory_relations as dr')
    ->leftJoin('directory as d', 'd.id', '=', 'dr.directory_id');

$base = DB::select(['dr.*', 'd.id as d_id', 'd.name', 'd.user_id'])
    ->from('directory as d')
    ->leftJoin('directory_relations as dr', 'd.id', '=', 'dr.directory_id')
    ->union($union);

DB::select('t1.*', 'v.hash')
    ->from(DB::raw("({$base->toSql()}) as t1"))
    ->mergeBindings($base->getQuery())
    ->leftJoin('videos as v', 't1.video_id', '=', 'v.id')
    ->where('t1.user_id', 265)
    ->get();