【问题标题】:A PHP Error was encountered Severity: Notice Message: Undefined variable: id Filename: models/singlequery_model.php Line Number: 4 [duplicate]遇到 PHP 错误 严重性:通知消息:未定义变量:id 文件名:models/singlequery_model.php 行号:4 [重复]
【发布时间】:2014-07-21 18:40:20
【问题描述】:
A PHP Error was encountered

Severity: Notice

Message: Undefined variable: id

Filename: models/singlequery_model.php

Line Number: 4

A Database Error Occurred

Error Number: 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 '' at line 1

SELECT * FROM 患者 WHERE id =

文件名:C:\xampp\htdocs\login\system\database\DB_driver.php

行号:330

这是查询单个结果的正确方法吗?

    <?php
class Singlequery_model extends CI_Model{
    function get_query(){
        $query = $this->db->query("SELECT * FROM patient WHERE id  = $id");

        if($query->num_rows() > 0 ){
            foreach ($query->result() as $row){
                $data[] = $row;
            }
            return $data;
        }
    }
}

下面是我的控制器

    <?php 
class Site extends CI_Controller{
    function __construct(){
      parent::__construct();
      $this->is_logged_in();
      $this->load->model('site_model');

    }
    function delete(){
      $this->site_model->delete_row();
      redirect('site/members_area');
    }

    function view(){

        $this->load->model('singlequery_model');

        $data['records'] = $this->singlequery_model->get_query();
        $this->load->view('patient_update', $data);

    }

}

【问题讨论】:

  • 你有什么问题?你有行 $query = $this-&gt;db-&gt;query("SELECT * FROM patient WHERE id = $id"); 和 Singlequery_model 你没有定义 $id

标签: php codeigniter


【解决方案1】:
$query = $this->db->query("SELECT * FROM patient WHERE id  = $id");

$id 没有在你的类中定义

为此,请确保使用$id 调用该函数:

$data['records'] = $this->singlequery_model->get_query($id);

我无法知道您的$id 来自哪里,所以您只能自己管理。

【讨论】:

  • 我在问什么是正确的查询方法,因为即使我把函数 get_query($id){ $query = $this->db->query("SELECT * FROM patient WHERE id = $id"); if($query->num_rows() > 0){ foreach ($query->result() as $row){ $data[] = $row; } 返回$数据; } } }
  • @JohnDanielVogue 请用您新的无效代码更新您的问题,我会相应地更新我的答案。但是只放function get_query($id)是不够的,你还必须把它添加到函数call
猜你喜欢
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 2015-12-06
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 2022-08-05
  • 2016-02-25
相关资源
最近更新 更多