<?php session_start(); if (!isset($_SESSON['newuser']) || !isset($_POST['submit'])) { ?> <html> <body> <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post"> Username: <input type="text" name="username"><br /> Password: <input type="password" name="password" value="********"<br /> <input type ="submit" name="submit" value="Submit!"> </form> </body> </html> <?php } else { $mysqluser = "root"; $mysqlpassword = "blablabla"; $host = "localhost"; $databasename = "lmfao"; $connection = mysql_connect($host, $mysqluser, $mysqlpassword); if (!$connecttion) { die("Could not connect" . mysql_error()); } else { mysql_select_db($databasename, $connection) or die('Could not connect' . mysql_error()); $newuser = mysql_real_escape_string($_POST['username']); $newpass = mysql_real_escape_string($_POST['password']); $query = "SELECT * FROM login WHERE username='$newuser' AND password='$newpass'"; $done = mysql_query($query); $count = mysql_num_rows($done); if ($count == 1) { session_register(newuser); echo "Login Success!"; } else { echo "You Failed to login"; } ?> PHP: Anybody got insight into whats wrong with the code? Im getting this error: Parse error: parse error in C:\Program Files\EasyPHP 3.0\www\design\login.php on line 41 but to my knowledge that line is fine.
You have forgotten to close 2 curly braces opened at the following line numbers: a) Line #19 b) Line #28