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].