dreamwest Posted December 30, 2007 Share Posted December 30, 2007 Im trying to filter hacking attempts on this form but keep getting this error message: Fatal error: Call to undefined function safe_ouput() in \get_post.php on line 14 What am I doing wrong?? <?php function safe_output($string) { $string = trim($string); $string = strip_tags($string); $string = htmlspecialchars($string); return $string; } if ($_GET ['action'] == 'send') { echo ('Data = '.safe_ouput($_GET['keyword'])); } ?> <form method="get" action ="get_post.php"> <fieldset> <legend>Programming Form</legend><br> <input type="text" name ="keyword" size ="25"><br><br> <input type="submit" name ="action" value ="send"> </fieldset> </form> Quote Link to comment https://forums.phpfreaks.com/topic/83696-solved-using-de-hacking-functions/ Share on other sites More sharing options...
PFMaBiSmAd Posted December 30, 2007 Share Posted December 30, 2007 You might want to run a spell checker on the code (you have got a spelling error.) Quote Link to comment https://forums.phpfreaks.com/topic/83696-solved-using-de-hacking-functions/#findComment-425843 Share on other sites More sharing options...
dreamwest Posted December 30, 2007 Author Share Posted December 30, 2007 Your right Thanks for the solution PFMaBiSmAd Quote Link to comment https://forums.phpfreaks.com/topic/83696-solved-using-de-hacking-functions/#findComment-425847 Share on other sites More sharing options...
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