Posted by peter on 08/28/07 15:17

> It returns null, which brings up a blank image.

then you can simply do something such as:-

<?PHP
$image = getRandomImage('LL/');
if (is_null($image))
{
$image = '';// Set to the image you wish
}
echo "<img src='$image' height=90 width=120 alt='LL'>";
?>

in the if statement place the path and name of the image you want to use as
the image that will be displayed if nothing is returned from the function.

[Back to original message]