如何使用選擇加入與指定laravel5

Question

我想從3表中Laravel5加入選擇任何字段，但對我來說

public function getJD($id){ 

    $result = []; 
    $result = JournalDetail::select('journal_detail.*, jdId as journal_detail, jdJ_id as journal_detail.journal_id .............,tranId as transactions_requiry,..............') 
     ->join('journal_requiry','journal_requiry.id','=','journal_detail.journal_id') 
     ->leftJoin('transactions_requiry','transactions_requiry.id','=','journal_requiry.tran_id') 
     ->where('journal_detail.journal_id','=',$id)->get(); 
    return $result; 

} 

錯誤它的第一時間，

QueryException in Connection.php line 624: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'as `as` from `tb_journal_detail` inner join `tb_journal_requiry` on `tb_journal_' at line 1 (SQL: select `tb_journal_detail`.* as `as` from `tb_journal_detail` inner join `tb_journal_requiry` on `tb_journal_requiry`.`id` = `tb_journal_detail`.`journal_id` left join `tb_transactions_requiry` on `tb_transactions_requiry`.`id` = `tb_journal_requiry`.`tran_id` where `tb_journal_detail`.`journal_id` = 26) 

但我不能這樣做，我不知道如何在Laravel5中做。 請幫助

Answer 1

把DB::raw(後選擇讓你的查詢將是：

$result = JournalDetail::select(DB::raw('journal_detail.*, jdId as journal_detail, jdJ_id as journal_detail.journal_id .............,tranId as transactions_requiry,..............')) 
     ->join('journal_requiry','journal_requiry.id','=','journal_detail.journal_id') 
     ->leftJoin('transactions_requiry','transactions_requiry.id','=','journal_requiry.tran_id') 
     ->where('journal_detail.journal_id','=',$id)->get();