PHPNewbie55 Posted December 20, 2007 Share Posted December 20, 2007 I started using <?php error_reporting(E_ALL); ?> At the beginning of all of my scripts... But now I can't do what I usually do without getting an error... If I start the script off like this with error_reporting(E_ALL); <?php error_reporting(E_ALL); if (!$action) { print "this"; } elseif ($action = "submit") { print "that"; } ?> I get an error message.. Notice: Undefined variable: action in /user/web/myurl.com/thefile.php on line 4 So how would I define this variable... because I thought that I was defining it by the if (!$action) and the elseif ($action = "submit") Link to comment https://forums.phpfreaks.com/topic/82459-error-reporting-question/ Share on other sites More sharing options...
~n[EO]n~ Posted December 20, 2007 Share Posted December 20, 2007 Define action at the top <?php error_reporting(E_ALL); // keep it blank or keep any value you want $action = ""; if (!$action) { print "this"; } elseif ($action = "submit") { print "that"; } ?> Link to comment https://forums.phpfreaks.com/topic/82459-error-reporting-question/#findComment-419227 Share on other sites More sharing options...
rajivgonsalves Posted December 20, 2007 Share Posted December 20, 2007 or your code should be <?php error_reporting(E_ALL); if (!isset($action)) { print "this"; } elseif (isset($action) && $action = "submit") { print "that"; } ?> Link to comment https://forums.phpfreaks.com/topic/82459-error-reporting-question/#findComment-419241 Share on other sites More sharing options...
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