php - ajax调用login.php，响应失败

Question

我正在使用ajax调用进行login.php..下面是我的ajax位代码和login.php代码..同时还提供了有效的用户名和密码..我收到了无效的用户名或密码消息..完全是无法理解这些代码行

$is_ajax = $_REQUEST['is_ajax'];
    if(isset($is_ajax) && $is_ajax) // in login.php, Many thanks

$("#login").click(function() {    
    var action = $("#form1").attr('action');
    var form_data = {
        username: $("#username").val(),
        password: $("#password").val(),
        is_ajax: 1
    };

    $.ajax({
        type: "POST",
        url: login.php,
        data: form_data,
        success: function(response)
        {
            if(response == 'success')
                $("#form1").slideUp('slow', function() {
                    $("#message").html("<p class='success'>You have logged in successfully!</p>");
                });
            else
                $("#message").html("<p class='error'>Invalid username and/or password.</p>");    
        }
    });

    return false;
});

这是login.php代码。

<?PHP
if(isset($_REQUEST['is_ajax']) && $_REQUEST['is_ajax']) {
    $is_ajax = $_REQUEST['is_ajax'];

    $uname = $_POST['username'];
    $pword = $_POST['password'];
    $uname = htmlspecialchars($uname);
    $pword = htmlspecialchars($pword);

    $user_name = "root";
    $pass_word = "root";
    $database = "test";
    $server = "127.0.0.1";

    $username = $_REQUEST['username'];
    $password = $_REQUEST['password'];

    $db_handle = mysql_connect($server, $user_name, $pass_word);
    $db_found = mysql_select_db($database, $db_handle);
    $uname = quote_smart($uname, $db_handle);
    $pword = quote_smart($pword, $db_handle);
    $SQL = "SELECT * FROM login WHERE L1 =  $uname AND L2 = md5($pword)";
    $result = mysql_query($SQL);
    $num_rows = mysql_num_rows($result);
    if ($num_rows > 0) {
        echo "success"; 
    }

    mysql_close($db_handle);
}
?>

Answer 1

一些事情:

1. 在查询中使用用户输入之前，您需要先对其进行清理。 htmlspecialchars 不这样做。使用mysql_real_escape_string(也许quote_smart可以做到这一点？)。
2. 您的 URL 未加引号，因此它被解释为代码。将 login.php 放在引号中。
3. 确保您了解 $_POST 和 $_REQUEST 是什么。看起来您实际上并没有在任何地方使用 $username 或 $password ，因此看起来这现在不会对您造成伤害，但您应该知道$_POST 数组的全部内容也可在 $_REQUEST 数组中找到(以及 $_GET 和 $_COOKIE 数组) >).
4. 为了进行调试，在 PHP 脚本中，您应该回显某种失败消息以查看脚本在哪里失败，而在 jQuery 脚本中，您应该将响应添加到失败消息中以查看实际响应是什么。<

试试这个:

$("#login").click(function() {    
    var action = $("#form1").attr('action');
    var form_data = {
        username: $("#username").val(),
        password: $("#password").val(),
        is_ajax: 1
    };

    $.ajax({
        type: "POST",
        url: "login.php",
        data: form_data,
        success: function(response)
        {
            if(response == 'success')
                $("#form1").slideUp('slow', function() {
                    $("#message").html("<p class='success'>You have logged in successfully!</p>");
                });
            else
                $("#message").html("<p class='error'>Invalid username and/or password. (debugging, response = '". response ."'</p>");    
        }
    });

    return false;
});

登录.php:

if(isset($_POST['is_ajax']) && $_POST['is_ajax'])
{
    $username = mysql_real_escape_string($_POST['username']);
    $password = mysql_real_escape_string($_POST['password']);

    $dbUsername = "root";
    $dbPassword = "root";
    $database = "test";
    $server = "127.0.0.1";

    $db_handle = mysql_connect($server, $dbUsername, $dbPassword);
    $db_found = mysql_select_db($database, $db_handle);
    $result = mysql_query("SELECT * FROM login WHERE L1 = '{$username}' AND L2 = MD5('{$password}')");
    $num_rows = mysql_num_rows($result);
    if ($num_rows > 0)
        echo "success";
    else
        echo "username '{$username}' and password '{$password}' not found";

    mysql_close($db_handle);
}
else
    echo "is_ajax = {$_POST["is_ajax"]}";