I am getting the error on line 26 as shown by my browser.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "tut";
$conn = mysqli_connect($servername, $username, $password, $database);
if (!$conn) {
die("Database connection failed: ".mysqli_connect_error());
}
if (isset($_POST['register']))
{
$user = $_POST['username'];
$pass = $_POST['password'];
$pass2=$_POST['password1'];
if(empty($username)||empty ($password)||empty($password1)){
echo "Oops! Can't leave any field blank";
}
elseif($pass!=$pass2){
echo "Passwords don't match";
}
else{
$phash = sha1(sha1($pass."salt")."salt");
$sql=IF NOT EXISTS (SELECT * FROM users WHERE username = '$user')
INSERT INTO users (id, username, password) VALUES ('', '$user', '$phash')
ELSE
RAISERROR 'Username exists, please select a different one';
$result = mysqli_query($conn, $sql);
}
}
?>
Is this not a correct way of writing the IF NOT EXISTS statement. Also when I try to execute this directly in XAMPP I get Unrecognised SQL statement error!