在 codeIgniter 中使用未定义的常量 roll_no - 假定为 'roll_no'答案

【问题标题】：Use of undefined constant roll_no - assumed 'roll_no' in codeIgniter在 codeIgniter 中使用未定义的常量 roll_no - 假定为 'roll_no'
【发布时间】：2016-11-20 08:41:46
【问题描述】：

我正在学习 CodeIgniter，遵循本教程：http://www.tutorialspoint.com/codeigniter/working_with_database.htm 面临以下错误：

遇到了 PHP 错误

严重性：通知

消息：使用未定义的常量 roll_no - 假定为“roll_no”

文件名：views/Stud_edit.php

行号：16

回溯：

文件：C:\xampp\htdocs\CI\CI-1\application\views\Stud_edit.php 线路：16 函数：_error_handler

文件：C:\xampp\htdocs\CI\CI-1\application\controllers\Stud_controller.php 线路：44 功能：查看

文件：C:\xampp\htdocs\CI\CI-1\index.php 线路：315 函数：require_once

这里是我的代码“Stud_edit.php”

<!DOCTYPE html> 

<html lang = "en">
<head> 
  <meta charset = "utf-8"> 
  <title>Students Example</title> 
  </head> 

  <body> 
  <form method = "" action = "">

     <?php 
        echo form_open('Stud_controller/update_student'); 
        echo form_hidden('old_roll_no',$old_roll_no); 
        echo form_label('Roll No.'); 
        echo form_input(array('id'=>'roll_no','name'=>'roll_no','value'=>$records[0]>roll_no));
        echo "<br/>"; 

        echo form_label('Name'); 
        echo form_input(array('id'=>'name','name'=>'name','value'=>$records[0]->name));
        echo "<br/>"; 

        echo form_submit(array('id'=>'submit','value'=>'Edit')); 
        echo form_close();
     ?> 

  </form>

Stud_controller.php

<?php 
  class Stud_controller extends CI_Controller {

  function __construct() { 
     parent::__construct(); 
     $this->load->helper('url'); 
     $this->load->database(); 
  } 

  public function index() { 
     $query = $this->db->get("stud"); 
     $data['records'] = $query->result(); 

     $this->load->helper('url'); 
     $this->load->view('Stud_view',$data); 
  } 

  public function add_student_view() { 
     $this->load->helper('form'); 
     $this->load->view('Stud_add'); 
  } 

  public function add_student() { 
     $this->load->model('Stud_Model');

     $data = array( 
        'roll_no' => $this->input->post('roll_no'), 
        'name' => $this->input->post('name') 
     ); 

     $this->Stud_Model->insert($data); 

     $query = $this->db->get("stud"); 
     $data['records'] = $query->result(); 
     $this->load->view('Stud_view',$data); 
  } 

  public function update_student_view() { 
     $this->load->helper('form'); 
     $roll_no = $this->uri->segment('3'); 
     $query = $this->db->get_where("stud",array("roll_no"=>$roll_no));
     $data['records'] = $query->result(); 
     $data['old_roll_no'] = $roll_no; 
     $this->load->view('Stud_edit',$data); 
  } 

  public function update_student(){ 
     $this->load->model('Stud_Model');

     $data = array( 
        'roll_no' => $this->input->post('roll_no'), 
        'name' => $this->input->post('name') 
     ); 

     $old_roll_no = $this->input->post('old_roll_no'); 
     $this->Stud_Model->update($data,$old_roll_no); 

     $query = $this->db->get("stud"); 
     $data['records'] = $query->result(); 
     $this->load->view('Stud_view',$data); 
  } 

  public function delete_student() { 
     $this->load->model('Stud_Model'); 
     $roll_no = $this->uri->segment('3'); 
     $this->Stud_Model->delete($roll_no); 

     $query = $this->db->get("stud"); 
     $data['records'] = $query->result(); 
     $this->load->view('Stud_view',$data); 
  } 
} ?>

【问题讨论】：

无法获取？它在哪里？
读取错误Line Number: 16
你是说，-> 而不是 => 在这一行？ echo form_input(array('id'=>'roll_no','name'=>'roll_no'));
$records[0]>roll_no
@u_mulder 谢谢，echo form_input(array('id'=>'roll_no','name'=>'roll_no','value'=>$records[0]->roll_no)) ;这有效

标签： php codeigniter

【解决方案1】：

你的语法有错误..

echo form_input(array('id'=>'roll_no','name'=>'roll_no','value'=>$records[0]>roll_no));

$records[0]>roll_no 中缺少 -..

echo form_input(array('id'=>'roll_no','name'=>'roll_no','value'=>$records[0]->roll_no));

【讨论】：