Php Login Page

[vexer007]

HELP PLEASE, I AM TRYING TO CREATE A LOGGIN IN PHP, BUT FOR SOME REASON, THE CODE IS JUST FRUSTRATION ME. HERE IS THE CODE BELOW

$query = ("SELECT count(username) as the_count FROM users WHERE username = '$username' AND email = '$email'");
$result = $conn->execute($query);
$state = ($result);
if($state == 0)
{
die("<META http-equiv=\"Refresh\" content=\"0;url=wrong.php\">");
}
else
setcookie("userName", $username);
die("<META http-equiv=\"Refresh\" content=\"0;url=index.php\">");
}
?>

THE PROBLEM IS THAT IT IS SENDING ME TO THE INDEX.PHP PAGE WHETHER OR NOT I INPUT THE WRONG INFO.

ANY HELP APPRECIATED.

[tekky]

what are you expecting this line to do?

$state = ($result);?

[vexer007]

To store the result of the query whether it return 1 for user exist or 0 for user dont exist.

or what do you reccommend?

[rgermain]

Good way to see what is going on with code is to see what your variables actually contain.

I would do an echo on the variables you are using and see what they are.

This will get you to a point as to where you can find what is wrong or least point into the right direction.

[vexer007]

I have echoed the SQL statement it shows the value of the variables i.e

SELECT username, pass FROM users WHERE username = 'zooghgh' AND pass = '12345625'

Thanks

[vexer007]

I HAVE EVEN TRIED:

                    $username = $_POST['username']; 
                    $password = $_POST['password']; 
                    $submit = $_POST['submit'];
                    if(isset($submit)) { 

            // SQL statement to build recordset.

$query = ("SELECT count(username) as the_count FROM users WHERE username = '$username' AND pass = '$password'");
$result = $conn->execute($query);

			if (!$result)
			{
			echo ("there were no match");
			}
			else {
			 echo ("There was a result");
			 }
			 }

IT STILL ALWAYS RETURNS THERE WAS A MATCH WHETHER OR NOT I INPUT THE WRONG DETAIL. PLEASE HELP...

[bpat1434]

If you're using a class, something will always be returned to $result. Try not using the class, and going about it as:
$result = mysql_query($sql);

Also, don't pull the username and password together, or if you do, don't make it an AND statement. Check after you pull it, just to make sure.

~Brett

[Cailean]

I'll just put how I would do it and maybe you can sort it out...

$username = $_POST['username'];
$password = $_POST['password'];
$submit = $_POST['submit'];

if(isset($submit)) {
	// SQL statement to build recordset.
	$query = ("SELECT username, pass FROM users WHERE username = '$username'");
	$result = my_sql_assoc($query);

if (!$result) {
	echo ("there were no match");
} elseif ($result['pass']===$password {
	echo ("There was a result");
} else {
	echo ("Password doesn't match username');
} 
} 

[vexer007]

Because it is Access DB, the usual MYSQl statements dont work.

This is the code so far, and this is the error I get:

$query =("SELECT TOP 1 username FROM users WHERE username = '$username'");
$result = $conn->execute($query);
$Recordset = ($result);
while (!$Recordset->EOF) {
$dbpassword = $Recordset->Fields("pass");
$Recordset->MoveNext();
}

 if($db_password == $password){ 
    echo ("Login successful, show them the page");
} 
else { 
 echo ("Login Failed, show login again, clear values ");
} 

}

ERROR:

Warning: (null)(): Invoke() failed: Exception occurred. Source: ADODB.Recordset Description: Item cannot be found in the collection corresponding to the requested name or ordinal. in e:\webareas\ou301\project\login.php on line 22
Login Failed, show login again, clear values

Thanks