php - 连接得到 4 个表，所有结果均按 id

Question

问题

现在我正在 codeigniter 中编辑表单，我想获取所有四个表数据。来自 Controller 的 event_id 和我想通过这个 id 获取数据。

型号

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

可能您需要字符串连接(否则变量无法正确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();
   }

调用未定义的方法 CI_DB_mysqli_driver::row_object() 错误
您没有选择表名称...因此您应该为您需要的表/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();