debug php code invoked by other php code
Posted: Sun Aug 16, 2009 4:12 am
How can I debug some php code which is invoked by other php files and not directly displayed in the html page ?
For example, this is the main php file:
But if I place echo "hello" inside showthumb.php, I don't see the output
thanks
For example, this is the main php file:
Code: Select all
...
echo "hello" //it works
$html .= '" target="_blank"><img src="'.$mosConfig_live_site.'/plugins/content/plugin_jw_sig/showthumb.php?img='.$_images_dir_.'/'.$images[$a]['filename'].'&width='.$_width_.'&height='.$_height_.'&quality='.$_quality_.'"></a></div></div>';
//above it calls showthumb.php with some parametersthanks