find out my error

Question

if (isset($_POST['login'])) {

$username = $_POST['uname'];
$password = $_POST['pass'];
}

$qry = "SELECT * FROM `admin` WHERE 'username'='$username' AND 'password'='$password';"

$run = mysqli_query($conn,$qry);

$row = mysqli_num_rows ($run);

if ($row <1)
{
?>

Answer 1

Alright, I think I get the issue now, you missed a semicolon at the end of the $qry variable declaration. Check if this will solve the problem.

$qry = "SELECT * FROM `admin` WHERE `username`='$username' AND `password`='$password';";

Hth, cmiiw

Answer 2

Can you elaborate more details on the problem you're dealing with? I suggest you to save the code in your profile and share the code link instead of writing it directly. I think you're only showing a part of it, what comes after the PHP code isn't available for analysis.

Answer 3

yes ofcource

Parse error: syntax error, unexpected '$run' (T_VARIABLE) in C:\xampp\htdocs\sms\login.php on line 52