Please help

[dudejma]

I get this error:

Parse error: syntax error, unexpected '}' in /home/virtu857/public_html/aandwtechsupport.co.cc/login_process.php on line 36

 

This is my code:

<?php
session_start();

include('connections/systems.php');

// Username and the password from form, and protected from MySQL injection
$email = mysql_real_escape_string(stripslashes($_POST['email']));
$password = mysql_real_escape_string(stripslashes($_POST['password']));
$result = mysql_query("SELECT * FROM users WHERE email='$email' and password='$password'");
if(mysql_num_rows($result) == 1) 
// If query returned 1 row{    
// Register nessecary sessions.    
session_register("email"); // It would be wise to do $_SESSION['email'] = $email;    
session_register("password");    

if ($level == "1")    
{       
     header("location: client.php");    
}    
else if ($level == "2")    
{        
     header("location: staff.php");    
}     
else if ($level == "3")    
{       
     header("location: techadmin.php");    
}    
else if ($level == "4")    
{         
     header("location: admin.php");    
}    
else    
{        
     header("location: login.php");    
}
}
?>

 

Line 36:

 }

 

Thanks for any help!

[hcdarkmage]

Your missing a beginning {:

if(mysql_num_rows($result) == 1) { //<--- Here

[radar]

if(mysql_num_rows($result) == 1)

 

no { after that..  try it and see if it works

 

hcdarkmage beat me to it ;(

[dudejma]

Thanks. Now there is a new error.

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/virtu857/public_html/aandwtechsupport.co.cc/login_process.php on line 10

 

Line 10:

if(mysql_num_rows($result) == 1){

[dudejma]

??? Any help, please!

[trq]

That means your query is failing. Considering 'password' is a mysql function you will likely need to change the name of that field to something else. or escape it using `backticks`.

[radar]

or you can do what i do is change your query line into 2 lines...

 

$sql = "SELECT * FROM users WHERE email='".$email."' AND password='".$password."'";

$result = mysql_query($sql);

 

but then commenting out the $result (or not even) and adding in: echo $sql; to verify that the query itself has the wanted information.