russia5 Posted August 12, 2006 Share Posted August 12, 2006 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.[code]<?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]?>?>[/code]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? Quote Link to comment Share on other sites More sharing options...
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