[RESOLVED] User Signup Problem

DiscoDrive

Hello there,

I have been dabbling with a user signup form, just to make sure I am familiar with the code. When it is working I will adapt it to fit my website design, but I'm having a problem with it at the moment.

The code below is giving me some problems because it gets as far as creating the query, but it will not execute it. If anybody can see the problem at a glance please let me know as it is causing me to tear my hair out. If you need any more info, I'll post anything up. Thankyou for your time.


<html><head><title>Adding a User</title></head><body>
<?php 
$self = $_SERVER['PHP_SELF'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$username = $_POST['username'];
$password = $_POST['password'];

if ( ( !$firstname ) or ( !$lastname )
					or ( !$username ) or ( !$password ) )
{

$form ="Please fill in the following:";
$form.="<form action=\"$self\"";
$form.=" method=\"POST\">First Name: ";
$form.="<input type=\"text\" name=\"firstname\"";
$form.=" value=\"$firstname\"> <br>Last Name: ";
$form.="<input type=\"text\" name=\"lastname\"";
$form.=" value=\"$lastname\"> <br>User Name: ";
$form.="<input type=\"text\" name=\"username\"";
$form.=" value=\"$username\"> <br>Password: ";
$form.="<input type=\"text\" name=\"password\"";
$form.=" value=\"$password\"><br>";
$form.="<input type=\"submit\" value=\"Submit\">";
$form.="</form>";
echo( $form ) ;
}

else
{ $conn = @mysql_connect( "localhost", "root", "******" )
or die("Could not connect to MySQL");
$db = @mysql_select_db( "noodlenews", $conn)
or die("Could not find database");
$sql = "insert into members
(firstname,lastname,username,password) values
(\"$firstname\",\"$lastname\",\"username\"
password(\$password\") )";
$result = @mysql_query( $sql, $conn)
or die("Could not execute query");
if( $result ) { echo ( "New user $username added" ); }
}
?>

YAOMK

You should not suppress errors with the @ character, but instead echo them out using mysql_error() so that you can debug your script.

Your query is incorrect, try this instead:

$pass = password($password) ; //not sure what this does .....
$sql = "INSERT INTO members (firstname,lastname,username,password) VALUES ('$firstname', '$lastname', '$username', '$pass' )";

DiscoDrive

Im sorry if im being slow, but im not sure i understand what you mean. If i place the mysql_error() command to show the error, where should i place it? Also, i tried the code you suggested but it still didnt work. Where was my query incorrect? Thankyou for your help non the less.

YAOMK

The backslashes you used to escape quotation marks in particular the one you used to (I guess) escape the dollar sign didn't make a lot of sense to me. The way you implement mysql_error() is as follows:

error_reporting(E_ALL); // this should go atop your script, or in a configuration file.
else
{ $conn = mysql_connect( "localhost", "root", "******" )
or die("Could not connect to MySQL: ".mysql_error());
$db = mysql_select_db( "noodlenews", $conn)
or die("Could not find database");
$pass = password($password) ; //not sure what this does ..... 
$sql = "INSERT INTO members (firstname,lastname,username,password) VALUES ('$firstname', '$lastname', '$username', '$pass' )";
$result = mysql_query( $sql, $conn)
or die("Could not execute query: ".mysql_error());
if( $result ) { echo ( "New user $username added" ); }
}

DiscoDrive

Ok then, I implemented the mysql_error() command as you said. The backslashes dont make a difference to the code, i tried getting rid of them and using ' instead of ", it makes no odds, so i left them their. When i change the password code:

$pass = password($password) ;

as you suggested, the script no longer works at all. The error I am getting when I test the code is as follows:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'password('jiofsd') )' at line 4

Im not sure what to do about this, I hope somebody can help me!! Much appreciated.

DiscoDrive

Ok problem solved, I simply changed the password section to:

$sql = "insert into members
(firstname,lastname,username,password) values
('$firstname','$lastname','username','password') )";

and it worked! Not sure why it didnt the first time around, and since i got this code from a book, Im also unsure how reliable it actually is, but the problem is solved now. Thanks to everyone who gave me any input.

YAOMK

I'm glad you solved your issue. I didn't know that password() was a mysql function , so of course it would need to be within your query as I now know what it does. However I doubt that this would have worked:

$sql = "insert into members
(firstname,lastname,username,password) values
(\"$firstname\",\"$lastname\",\"username\"
password(\$password\") )";

this:

$sql = "insert into members
(firstname,lastname,username,password) values
(\"$firstname\",\"$lastname\",\"username\"
password(\"$password\"))";

or the following would have worked, but not your original query hence your original issue.

$sql = "insert into members
(firstname,lastname,username,password) values
(\"$firstname\",\"$lastname\",\"username\"
password('$password'))";

Take care.