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Debugging with echo


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#1 russia5

russia5
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Posted 12 August 2006 - 10:06 PM

I have a php problem and I read that I can debugg my script calling echo for all of he commands. I am not sure what I am suppose to see.  Here is an example of what I did and I got know results. (in bold at the end of the script and the variable is about three lines down.

<?php 
	$command = $imagemagik_path.'convert -resize ';

	// resize for thumbnail 1 size
	$current_command = $command.$thumb_size1.' "'.$old.'" '.$thumb1;
	exec($current_command, $retarray, $retval);
	// echo $current_command;
	if ($retval != 0) // error
	{
		echo 'error upon resizing for thumbnail 1 : '.$old.' to '.$thumb1;
	}

	// resize for thumbnail 2 size
	$current_command = $command.$thumb_size2.' "'.$old.'" '.$thumb2;
	exec($current_command, $retarray, $retval);
	// echo $current_command;
	if ($retval != 0) // error
	{
		// echo 'error upon resizing for thumbnail 2';
	}
	
	[b]echo '$current_command';[/b]?>

?>

Also, can anyone explain what the $command.$thumb_size1  means?  I know $command is the path to imagemagick.  But what would $thumb_size1 mean and does it need declared someplace within the directory it is included in? 




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