pgrevents Posted May 23, 2009 Share Posted May 23, 2009 Its simple i know but i cannot see the errors of my ways. I am creating a sign up form and I am passing the details via ajax. I know that both variables are sending and getting picked up by the script but its returning false info. Basicly its a validation to check if the user has confirmed their password correctly. here is the code <?php $pass = $_POST['pword']; $check = $_POST['check_pass']; if ($pass == $check) { echo "yes"; } else { echo "no"; } ?> i been staring at this machine all day thats prob why i cant see the error :0 any help?? Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/ Share on other sites More sharing options...
jackpf Posted May 23, 2009 Share Posted May 23, 2009 Try this echo $pass.' - '.$check; And see what's output. Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/#findComment-840904 Share on other sites More sharing options...
razta Posted May 23, 2009 Share Posted May 23, 2009 if ($pass === $check) Maybe that will work? Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/#findComment-840905 Share on other sites More sharing options...
RussellReal Posted May 23, 2009 Share Posted May 23, 2009 <?php $pass = $_POST['pword']; $check = $_POST['check_pass']; if (($pass == $check) && (strlen($pass)) { echo "yes"; } else { echo "no"; } ?> this would be a tad more secure, however, there is no error in ur php.. I think you're sending them as the wrong names you're expecting Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/#findComment-840906 Share on other sites More sharing options...
pgrevents Posted May 23, 2009 Author Share Posted May 23, 2009 Thanks for the hints. I tried every if expresion i could think of and not working. I am sending I know they are getting recieved correctly ill try some fidgeting Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/#findComment-840908 Share on other sites More sharing options...
jackpf Posted May 23, 2009 Share Posted May 23, 2009 Try this echo $pass.' - '.$check; And see what's output. What did that output then? Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/#findComment-840910 Share on other sites More sharing options...
pgrevents Posted May 23, 2009 Author Share Posted May 23, 2009 that was the first thing i done when it wasnt working its the same variables as passed Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/#findComment-840915 Share on other sites More sharing options...
pgrevents Posted May 23, 2009 Author Share Posted May 23, 2009 solved it was the ajax a letter out of place thanks for the help Link to comment https://forums.phpfreaks.com/topic/159413-simple-if/#findComment-840917 Share on other sites More sharing options...
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