Warnign in a foreach loop

Question

I can't get my data from database , I keep getting the same error

Notice: Undefined variable: articles in :\xampp\htdocs\SalesManagementProject\Views\articles.php on line 25
Warning: Invalid argument supplied for foreach() in :\xampp\htdocs\SalesManagementProject\Views\articles.php on line 25

my loop

 <?php
 foreach($articles as $art)
 {
  echo '<tr>
         <td>'.$art['Id'].'</td>
         <td>'.$art['ref'].'</td>
         <td>'.$art['desig'].'</td>
        </tr>';
 }
 ?>

Model

 <?php
class ArticlesModel
{
    public function getArticles()
    {
        $pdo=new PDO_connect();
        $q= $pdo->PDO()->query('SELECT * FROM articles');
        return $q->fetchAll();
    }
}
?>

Controller

<?php
class ArticlesControllers
{
    public function loadArticles()
    {
        include (dirname(__FILE__).'/../Models/ArticlesModel.php');
        $art_table = new ArticlesModel();
        $articles = $art_table->getArticles();
        include(dirname(__FILE__).'/../Views/articles.php');
    }
}
?>

index

<?php
   session_start();
  include ('PDO_connect.php');
   $p='home';

  if (isset($_GET['p']))
  {
       $p=$_GET['p'];
   }
   include('Controllers/ArticlesControllers.php');
$arts=new ArticlesControllers();
$arts->loadArticles();
?>

So what i must do?

where is `$articles` defined? find that first – Kevin Aug 29 '16 at 02:09 — Kevin, Aug 29 '16 at 02:09
which one is your line 25 clarify – Mirza Obaid Aug 29 '16 at 04:55 — Mirza Obaid, Aug 29 '16 at 04:55
line 25 the foreach line Mirza Obaid – M. Hasimi Aug 29 '16 at 12:21 — M. Hasimi, Aug 29 '16 at 12:21
articles my table in database – M. Hasimi Aug 29 '16 at 12:25 — M. Hasimi, Aug 29 '16 at 12:25

Forbs · Answer 1 · 2016-08-29T04:04:14.900

-1

Simple, in your controller section after the public function loadArticles() { add

global $articles;

Because $articles is assigned inside of a function, you need to global it to read it in another area.
BTW you should define $articles as an array in index.php to make it properly defined

 $articles = array();

BTW passing a named variable in / out of a class is not very proper programming, but I wanted you to have a quick solution.

A better solution is to have the public function return ($articles) and in your index.php $articles = $arts->loadArticles();

edited Aug 29 '16 at 04:04

answered Aug 29 '16 at 03:58

Forbs

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Thank you I did that but its still now work ok there is no Warning now but does not bring anything from database :( – M. Hasimi Aug 29 '16 at 12:20
Well that's a different matter you will have to use `print_r` in various places – Forbs Aug 29 '16 at 14:09

Warnign in a foreach loop

1 Answers1