This is on the site
<form action="<?php echo URL; ?>inventory/connect" method="post">
<label>Username</label>
<input type="text" name="username" required />
<label>Password</label>
<input type="password" name="password" required />
</select>
<input type="submit"/>
</form>
this is my connect.php file
[code]$server = ‘localhost’;
$db = ‘cars’;
$user = $_POST[“username”];
$pass = $_POST[“password”];
$mysqli = new mysqli($server, $user, $pass, $db);
$mysqli_report(MYSQLI_REPORT_ERROR);[/code]
This is my error
[code]Warning: mysqli::mysqli(): (HY000/1045): Access denied for user ‘hotshots’@‘localhost’ (using password: YES) in D:\EasyPHP\data\localweb\application\views\inventory\connect.php on line 7
Notice: Undefined variable: mysqli_report in D:\EasyPHP\data\localweb\application\views\inventory\connect.php on line 8
Fatal error: Function name must be a string in D:\EasyPHP\data\localweb\application\views\inventory\connect.php on line 8[/code]
Im trying to make it so they can log in into the database with what they type in the website and if it is correct it will continue otherwise it will return to same page with a error