我使用php和ajax创建了一个登录脚本。密码我回显值,作为回报我发现文本框值和数据库值相同,但仍然跳转到else语句。我无法弄清楚代码的问题或我做错了什么。可能是其中一个人可以看到我的代码中是否有任何错误
$username = $_POST['username'];
$pass = mysqli_real_escape_string($connection, $_POST['password']);
$pass = md5($pass);
$check = mysqli_query($connection, "SELECT * FROM users WHERE username = '$username'");
$result = mysqli_num_rows($check);
if($result == 1) {
$data = mysqli_fetch_array($check);
$password_check = $data[2];
if($password_check == $pass) {
$_SESSION['uid'] = $data[0];
$_SESSION['name'] = $data[3];
echo "Success";
} else {
echo "<div class='message'>Invalid Password</div>";
}
} else {
echo "<div class='message'>Username/Password did not matched</div>";
}