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Home/ Questions/Q 756845
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Editorial Team
  • 0
Editorial Team
Asked: May 14, 20262026-05-14T15:17:31+00:00 2026-05-14T15:17:31+00:00

This should be very basic, but I am a little stumped! Here is my

  • 0

This should be very basic, but I am a little stumped!

Here is my array:

$menu = array(
  'Home',

  'Stuff'=>array(
    'Losta Stuff',
    'Less Stuff',
    'Ur moms stuff',
    'FAQ'
 ),
  'Public Works'

);

Here is my logic:

echo "<ol>\n";
foreach( (array)$menu as $header )
{
  echo '  <li><b>'.$header."</b><br />\n";
 echo '  <ol>';
  foreach( (array)$header as $headers )
  {
    echo '    <li>'.$headers.".</li>\n";
  }
  echo '  </ol>';
}
echo "</ol>\n";

As you can see, Home and Public Works don’t have data in the them, so I get a

Warning: Invalid argument supplied for foreach() in test.php on line ##

If I add (array) to $header like this: foreach( (array)$header as $headers ), It no longer gives me the error, but it just displays the $header as the $headers (i.e. Home – Home, Instead of Home – nothing).

Basically, if the data is empty, I want it to do nothing!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T15:17:32+00:00Added an answer on May 14, 2026 at 3:17 pm

    You should test for whether or not the current item that you’re trying to echo is an array, which can be done with is_array, and then act accordingly. Something like the following might do the trick.

    <?php 
    
    $menu = array(
        'Home',
        'Stuff'=>array(
            'Losta Stuff',
            'Less Stuff',
            'Ur moms stuff',
            'FAQ'
        ),
        'Public Works'
    );
    
    echo "<ol>\n";
    foreach($menu as $menuName => $header )
    {
        if (!is_array($header))
        {
            echo '  <li><b>'.$header."</b><br />\n";
        }
        else
        {
            echo "<li><b>$menuName</b><ol>";
            foreach($header as $headers )
            {
                echo '    <li>'.$headers.".</li>\n";
            }
            echo "</ol></li>";
        }
    }
    echo "</ol>\n";
    
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