Signup Code Not Working

jwinchester1

Hello again everybody!

I'm trying to setup a new users signup page which inserts UserName, Password, EmailAddress, and LastName into table Users. The form displays correctly, but when submit is clicked it just refreshes the signup page... even if NOTHING is entered. Why aren't the checks below working? No error message is displayed, and nothing is entered into the db? Any suggestions?


<table border='0' width='50%' cellspacing='0' cellpadding='0' align=center>
<form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
<input type=hidden name="todo" value="post">

<tr bgcolor='#f1f1f1'>
<td align=center colspan=2><font face='Verdana' Size='2'><b>Signup</b></td></tr>
<tr ><td >&nbsp;<font face='Verdana' size='2' >User ID</td><td ><font face='Verdana' 

size='2'><input type=text name=UserName></td></tr>

<tr bgcolor='#f1f1f1'><td >&nbsp;<font face='Verdana' size='2' >Password</td><td ><font 

face='Verdana' size='2'><input type=text name=Password></td></tr>
<tr ><td >&nbsp;<font face='Verdana' size='2' >Re-enter Password</td><td ><font 

face='Verdana' size='2'><input type=text name=Password2></td></tr>


<tr bgcolor='#f1f1f1'><td ><font face='Verdana' size='2' >&nbsp;Email</td><td  ><input 

type=text name=EmailAddress></td></tr>
<tr ><td >&nbsp;<font face='Verdana' size='2' >Last Name</td><td ><font face='Verdana' 

size='2'><input type=text name=LastName></td></tr>

<tr ><td >&nbsp;<font face='Verdana' size='2' >I agree to terms and conditions</td><td 

<font face='Verdana' size='2'><input type=checkbox name=agree value='yes'></td></tr>

<tr bgcolor='#f1f1f1'><td align=center colspan=2><input type=submit value=Signup></td></tr>
</table>

<center>
<br><br><font face='Verdana' size='2' ><a href='login.php'>Already have a Username and 

Password? Please Login</a></font></center> 

<?php

if(isset($todo) and $todo=="post"){

$status = "OK";
$msg="";

if(!isset($UserName) or strlen($UserName) <1){
$msg=$msg."UserName MUST be filled in.<BR>";
$status= "NOTOK";}	

if(mysql_num_rows(mysql_query("SELECT UserName FROM Users WHERE UserName = '$UserName'"))){
$msg=$msg."UserName already exists. Please try another one<BR>";
$status= "NOTOK";}					


if ( strlen($Password) < 3 ){
$msg=$msg."Password must be more than 3 char legth<BR>";
$status= "NOTOK";}					

if ( $Password <> $Password2 ){
$msg=$msg."Both passwords are not matching<BR>";
$status= "NOTOK";}					


if ($agree<>"yes") {
$msg=$msg."You must agree to terms and conditions<BR>";
$status= "NOTOK";}	

if($status<>"OK"){ 
echo "<font face='Verdana' size='2' color=red>$msg</font><br><input type='button' 

value='Retry' onClick='history.go(-1)'>";
}else{ // if all validations are passed.
$query=mysql_query("insert into Users(UserID,Password,EmailAddress,LastName) 

values('$UserName','$Password','$EmailAddress','$LastName')");
echo "<font face='Verdana' size='2' color=green>Welcome, You have successfully signed 

up<br><br><a href=login.php>Click here to login</a><br></font>";
}
}

Thanks,

Pehape

you use POST method, so use $POST['urvarname'] to get your variable value...
like this one:
if(!isset($POST['UserName']) or strlen($POST['UserName']))
or yu could fill it in ur variable first..
$user=$POST['username'];
so...it will be:

if(!isset($User) or strlen($User)

apply it in all your post variable..

Crowly

Nothing to do with your problem, but if you want to follow the W3C standards <form> needs to come before <table>, like this:

<form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
<table border='0' width='50%' cellspacing='0' cellpadding='0' align=center>