加入获取4个表全部由ID结果

Question

问题

现在，我正在在CodeIgniter中进行编辑表单，我想获取所有四个表数据。来自控制器来自Event_ID，通过此ID，我想获取数据。

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

可能是您需要字符串串联（outhewise变量未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();
   }

呼叫未定义的方法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();