mysql - 如何使用select连接并在laravel5中指定任何字段

我想从3个表中选择一个加入Laravel5的表，但这是我第一次

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;

}

错误，

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)

但我不能这么做，我不知道怎么在拉瓦夫勒5。
请帮忙

最佳答案

在选择后放置DB::raw(，这样您的查询将是：

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

关于mysql - 如何使用select连接并在laravel5中指定任何字段，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/35264955/