连接get 4个表，所有结果均为id

Question

问题

现在，我正在代码点火器中编辑表单，我想获得所有四个表数据。从控制器中来的是event_id，通过这个id，我想获得数据。

模型

public function get_list_for_edit($event_id){
    return

   $this->db->select('*')
            ->from('events e')
            ->join('events_location el','el.events_id =' $event_id,'left')
            ->join('events_photos ep',
                    'ep.events_id =' $event_id,'left')
            ->join('push_notifications pn',
                    'pn.events_id =' $event_id,'left')
            ->where('e.event_id =' $event_id)
            ->row_object();
 }

错误

消息:语法错误，意外的'$event_id‘(T_VARIABLE)

Answer 1

可以是您需要字符串连接(当然，变量in没有正确地添加到字符串的其余部分)

  public function get_list_for_edit($event_id){
      return

     $this->db->select('*')
              ->from('events e')
              ->join('events_location el','el.events_id =' . $event_id,'left')
              ->join('events_photos ep',
                      'ep.events_id =' . $event_id,'left')
              ->join('push_notifications pn',
                      'pn.events_id ='. $event_id,'left')
              ->where('e.event_id =' . $event_id)
              ->row_object();
   }

用于调用未定义方法CI_DB_mysqli_driver::row_object()错误

你不能选择表名..。因此，您应该为所需的表/模型添加get('TABLE_NAME')方法

     $this->db->select('*')
              ->from('events e')
              ->join('events_location el','el.events_id =' . $event_id,'left')
              ->join('events_photos ep',
                      'ep.events_id =' . $event_id,'left')
              ->join('push_notifications pn',
                      'pn.events_id ='. $event_id,'left')
              ->where('e.event_id =' . $event_id)->get('TABLE_NAME')
              ->row_object();