Question

CI中有这个错误：

A PHP Error was encountered

Severity: Notice

Message: Undefined variable: news

Filename: controllers/First.php

Line Number: 33

A PHP Error was encountered



Severity: Notice

Message: Undefined variable: news

Filename: views/welcome_view.php

Line Number: 1

A PHP Error was encountered

Severity: Warning

Message: Invalid argument supplied for foreach()

Filename: views/welcome_view.php

Line Number: 1

有我的控制器：

<?php
if ( ! defined('BASEPATH')) exit ('No direct script access allowed');

   class First extends CI_Controller
   {

    public function __construct()
    {
    parent::__construct();
    $this->load->model('materials_model');
    }    


    public function index()
      {
        $this->load->view('header_view');
        $this->load->view('menu_view');
        $this->load->view('about_me_view');
        $this->load->view('navigation_view');
        $this->load->view('search_view');
        $this->load->view('main_text_view');
        $this->load->view('footer_view');

      }


    public function mat()
    {
       $this->load->model('materials_model');
       $this->materials_model->get();
       $data['news'] = $this->materials_model->get();

       $this->load->view('welcome_view',$news);

       if(empty($data['news']))
       {
        echo'Array is Null';
       }

       else
       {
        echo'Array has info';
       }
    }

我的模特：

<?php
if ( ! defined('BASEPATH')) exit ('No direct script access allowed');

class Materials_model extends CI_Model
{
    public function get()
    {
        $query = $this->db->get('materials');
        return $query->result_array();
    }


}

?>

我的观点：

<?php foreach ($news as $one):?>
<?=$one['author']?>
<?php endforeach; ?>

所以我知道，我传入视图的数组是非NULL（我检查它是否为print_r和if，否则构造），但我可以传递它并查看它。我有什么错误？

Answer 1

$this->load->view('welcome_view',$news);

永远不会定义

$news。也许你的意思是使用$data ?, CI视图希望将一个关联数组传递给它们。

Answer 2

设置变量：

$data['news'] = $this->materials_model->get();

将其传递给视图：

$this->load->view('welcome_view', isset($data) ? $data : NULL);

在视图中作为结果数组访问它：

<?php 
    foreach($news as $item)
    {
        echo $item['id']." - ".$item['some_other_column_title']
    } 
?>

作为一个选项，您可以在模型中更改get（）方法以返回对象，而不是数组。

public function get()
{
    $query = $this->db->get('materials');
    return $query->result();
}

然后在视图中你需要处理$ news作为对象，如：

<?php 
    foreach($news as $item)
    {
        echo $item->id." - ".$item->some_other_column_title
    } 
?>

Answer 3

在Controller中，您需要传递数据而非

之类的新闻

$data['news'] = 'Something';
$this->load->view('welcome_view',$data);

在您的视图中，您可以使用

foreach($news as $new)
{
   //Go Here
}

codeigniter中的未定义变量

3 个答案: