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> 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.) 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 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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