Registration form error.

[Foser]

Here's my script.

<?php
include("config.php");
//user data
$username = mysql_real_escape_string($_POST['username']);
$email = mysql_real_escape_string($_POST['mail']);
$password = md5($_POST['pw']);
$rights = "user";

$taken = false;

$q = "SELECT * FROM user_info";
$ms = mysql_query($q);

while($user_info = mysql_fetch_assoc($ms)){
if ($_POST['user'] == $userinfo['username']) {
$taken = true;
echo "This username has already been taken.<br>";}

if ($_POST['email'] == $userinfo['e-mail']){
$taken = true;
echo "This E-mail has already been registered.<br>.";}} 


if (!$take){
$qu = "INSERT INTO user_info (username, email, password, rights) VALUES('$username', '$email', '$password', '$rights')";
mysql_query($qu) or die mysql_error();
echo "You are now registered, you may now login."; }

?>

 

the error is on the line:

$qu = "INSERT INTO user_info (username, email, password, rights) VALUES('$username', '$email', '$password', '$rights')";

 

 

I can't seem to see any error, is it possible that my data type is set incorectly for the data?

Their all set to varchar.

[mmarif4u]

What specific error u got, pls post the error.

[Foser]

Parse error: syntax error, unexpected T_STRING in C:\WAMP\www\Tutorials\PHP_MYSQL\Simple_MySQL\Login\regproc.php on line 26

 

Thanks for responding

[mmarif4u]

This is:

if (!$take)

 

OR:

if (!$taken)

[Foser]

thanks for the correction for that part but it still has the same error, so that wasin't the main cause of the problem

[mmarif4u]

May be this will help you.

 

while($user_info = mysql_fetch_assoc($ms)){
if ($_POST['user'] == $user_info['username']) {
$taken = true;
echo "This username has already been taken.<br>";}

if ($_POST['email'] == $user_info['e-mail']){
$taken = true;
echo "This E-mail has already been registered.<br>.";}}

[Foser]

That does not work, and to me it looks exacly the same as mine... But I still copy pasted it and no it did not work.  ???

[Foser]

bump

[chocopi]

For mine i use this, yours seems abit over complicated

 

<?php

$username = mysql_real_escape_string($_POST['username']);
$email = mysql_real_escape_string($_POST['email']);

$check_user = mysql_query("SELECT username FROM table WHERE username='$username'"); 
$username_exist = mysql_num_rows($check_user);

$check_email = mysql_query("SELECT email FROM table WHERE email='$email'"); 
$email_exist = mysql_num_rows($check_email);

// form

if ($username_exist > 0)
{
	echo ("Sorry, but your username already exists. Please select another. <br />");
}

if ($email_exist > 0)
{
	echo ("Sorry, but you may only have one account per email address. <br />");
}
?>

 

Hope that helps,

 

~ Chocopi

[Foser]

<?php
include("config.php");
//user data
$username = mysql_real_escape_string($_POST['username']);
$email = mysql_real_escape_string($_POST['mail']);
$password = md5(sha1(md5(md5($_POST['pw']))));
$rights = "user";
// this line is line 8
$check_user2 = mysql_query("SELECT user FROM user_info WHERE user = '$username'");
$check_user = mysql_num_rows($check_user2);
$check_email2 = mysql_query("SELECT email FROM user_info WHERE email = '$email'");
$check_email = mysql_num_rows($check_email2);


if ($username > 0 ) {
echo "This username has already been taken.<br>";}

if ($email > 0 ){
echo "This E-mail has already been registered.<br>.";}


else {
$qu = "INSERT INTO user_info (username, email, password, rights) VALUES('$username', '$email', '$password', '$rights')";
mysql_query($qu) or die(mysql_error());
echo "You are now registered, you may now login."; }

?>

 

My error I have now is very odd. it is at a line where no information or commands are.

Parse error: syntax error, unexpected T_LOGICAL_OR in C:\WAMP\www\Tutorials\PHP_MYSQL\Simple_MySQL\Login\config.php on line 8

 

I amde it easy to spot on my code.

[DJTim666]

LMFAO. If you read that error very closly it will tell you that you have an error in your config.php file. ! So check line 8 in ur config.php file, and that will solve your error.

[Foser]

aha, ok i fixed it, now i it works

 

but get a

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\WAMP\www\Tutorials\PHP_MYSQL\Simple_MySQL\Login\regproc.php on line 10

  error, What is another similar command I can use to replace it with so I dont get this error.

 

thanks alot guys!

[mmarif4u]

Try this instead of Mysql_num_rows use mysql_result.

<?php
include("config.php");
//user data
$username = mysql_real_escape_string($_POST['username']);
$email = mysql_real_escape_string($_POST['mail']);
$password = md5(sha1(md5(md5($_POST['pw']))));
$rights = "user";
// this line is line 8
$check_user2 = mysql_query("SELECT user FROM user_info WHERE user = '$username'");
$check_user = mysql_result($check_user2);
$check_email2 = mysql_query("SELECT email FROM user_info WHERE email = '$email'");
$check_email = mysql_result($check_email2);


if (($check_user, 0) > 0 ) {
echo "This username has already been taken.<br>";}

if (($check_email,0) > 0 ){
echo "This E-mail has already been registered.<br>.";}


else {
$qu = "INSERT INTO user_info (username, email, password, rights) VALUES('$username', '$email', '$password', '$rights')";
mysql_query($qu) or die(mysql_error());
echo "You are now registered, you may now login."; }

?>

[Foser]

Warning: Wrong parameter count for mysql_result() in C:\WAMP\www\Tutorials\PHP_MYSQL\Simple_MySQL\Login\regproc.php on line 10

Warning: Wrong parameter count for mysql_result() in C:\WAMP\www\Tutorials\PHP_MYSQL\Simple_MySQL\Login\regproc.php on line 12
You are now registered, you may now login.

 

Same issue, maybe is it because wamp does not support these functions?