How to check if my foreach loop has a bug? - php

Question

I was just quickly throwing together a script for an article website where I retrieve articles out of a database.
Here is my index.php script
<?php
// include header
require("include/header.php");
require("include/helperfunctions.inc.php");
?>
<!-- page content -->
<!-- left section -->
<section id="contentleft">
<?php require("include/functions.php");
displayArticles();
foreach ($articles as $article) : ;
?>
<h2>Recent Articles</h2>
<ul>
<li><?php echo htmlout($articles['id']) ; ?></li>
<li><?php echo htmlout($articles['title']) ; ?></li>
<li><?php echo htmlout($articles['summary']) ; ?></li>
</ul>
<?php endforeach; ?>
</section>
<!-- right content -->
<section id="contentright">
</section>
<?php
// include footer
require("include/footer.php");
?>
Here is the start of the function library
function displayArticles($order="publicationdate DESC"){
// connect to the database
include("include/db.inc.php");
$query = "SELECT id, title, summary FROM maths order by ". $order . " limit 10";
// query the database
$result = mysqli_query($link, $query);
// error checking
if(mysqli_connect_errno()){
// $error = "error fetching articles";
echo " could not connect: " . mysqli_connect_errno() . " ";
exit();
}
// loop through storing values into array
while($row = mysqli_fetch_array($result)){
$articles[] = array('id'=>$row['id'] , 'title'=>$row['title'],'summary'=>$row['summary']);
}
}
?>
I get this error:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\Apache24\htdocs\include\functions.php on line 17 Recent Articles Notice: Undefined variable: articles in C:\Apache24\htdocs\home.php on line 14 Warning: Invalid argument supplied for foreach() in C:\Apache24\htdocs\home.php on line 14

Answer 1

It looks like a scope issue with the $articles object. You create $articles in the displayArticles() function.. but outside of the function, your other code does not know about it. Try returning $articles inside displayArticles(), and replace your displayArticles(); call at the top of the first page with:
$articles = displayArticles();
Also, as panique in the comments pointed out, you are referencing the wrong object inside your foreach block. Remove the 's' at the end of each $articles[blah], so that it reads $article[blah].