在我的登录脚本中一切正常。我在我的 div (id=login_reply) 中得到了正确的响应， session 开始了。但是每当我刷新页面时，login_reply 就消失了。我怎样才能保留 login_reply？谢谢!

这是 php:

if (isset($_POST['username'], $_POST['password']))
{
    $username = mysql_real_escape_string($_POST['username']);
    $password = mysql_real_escape_string(md5($_POST['password']));

    $check = mysql_query("SELECT * FROM `userbase` WHERE `user_name` = '$username'");
    if (mysql_num_rows($check) > 0)
    {
        while ($row = mysql_fetch_assoc($check))
        {
            $user_id = $row['user_id'];
            $user_name = $row['user_name'];
            $user_email = $row['user_email'];
            $user_password = $row['user_password'];

            if ($password == $user_password)
            {
                $_SESSION['user_id'] = $user_id;
                if (isset($_SESSION['user_id']) && $_SESSION['user_id'] != '')  
                    echo "Welcome back '$user_name'!";
                else
                {
                    echo 'no';
                }

            }
            else
                echo 'no';
        }
    }
    else
        echo 'no';  
}

这是jQuery

$(document).ready(function()
{

$('#login').click(function()
{
    var username = $('#username').val();
    var password = $('#password').val();

    $.ajax(
    {
        type: 'POST',
        url: 'php/login.php',
        data: 'username=' +username + '&password=' + password,
        success: function(data)
        {
            if (data != 'no')
            {
                $('#logform').slideUp(function()
                {
                    $('#login_reply').html(data).css('color', 'white');
                });
            }
            else                    
                $('#login_reply').html('Invalid username or password').css('color', 'red');
        }
    });
    return false;
});
});

最佳答案

问题是 JS 只是客户端脚本语言——它只在客户端浏览器上处理。

如果您想使用 AJAX 登录，可以，但是您必须将登录用户的值存储到 session (或 cookie)中。然后在每次页面加载时，您将检查该 session 或 cookie 是否已设置并存在，如果存在，您将在登录后记下与 jQuery 对应的 HTML...

换句话说，这就是你要做的:

1. 页面加载 - 用户未登录
2. 用户填写登录凭据并点击“登录”
3. 检查该用户是否存在，如果存在，则保存 $_SESSION['username'](例如)
4. 在 AJAX 中，您填写 $('#login_reply')
5. 用户点击您网站内的某个链接
6. 您检查您是否在 $_SESSION['username'] 中有一个值(或设置了索引)
7. 如果是，通过 PHP 您将填充 #login_reply div，如果不是，您将显示登录表单...

希望这有助于...

EDIT1:您还应该以客户端浏览器上没有 JS 工作(禁用)的方式实现登录功能，因此应该考虑正常的 POST...

EDIT2:我还想指出一些一般的编程错误......

这是您添加了我的评论的代码:

if (isset($_POST['username'], $_POST['password']))
{ // <-- This is a .NET style of writing the brackets that I don't like much and I guess PHP don't like .NET either :-)
    $username = mysql_real_escape_string($_POST['username']);
    $password = mysql_real_escape_string(md5($_POST['password']));

    $check = mysql_query("SELECT * FROM `userbase` WHERE `user_name` = '$username'"); // <-- mysql_* method calls should be replaced by PDO or at least mysqli
    // Also the query could be improved
    if (mysql_num_rows($check) > 0)
    {
        while ($row = mysql_fetch_assoc($check)) // <-- why calling while when You expect ONLY one user to be found?
        {
            $user_id = $row['user_id'];
            $user_name = $row['user_name'];
            $user_email = $row['user_email'];
            $user_password = $row['user_password']; // <-- What are these last 4 lines good for? This is useless...

            if ($password == $user_password) // <-- this condition SHOULD be within the SQL query...
            {
                $_SESSION['user_id'] = $user_id;
                if (isset($_SESSION['user_id']) && $_SESSION['user_id'] != '')  // <-- this condition is useless as You have just set the session variable... 
                // ALSO, if You use brackets with else it is good to use brackets also with if
                    echo "Welcome back '$user_name'!";
                else
                {
                    echo 'no';
                }

            }
            else
                echo 'no';
        }
    }
    else
        echo 'no';  
}

这是我重写的代码(仍然使用 mysql_*，抱歉):

if (isset($_POST['username'], $_POST['password'])) {
    $username = mysql_real_escape_string($_POST['username']);
    $password = mysql_real_escape_string(md5($_POST['password']));

    $check = mysql_query("SELECT * FROM `userbase` WHERE `user_name` = '{$username}' AND `user_password` = '{$password}' LIMIT 1"); // <-- We check whether a user with given username AND password exists and we ONLY want to return ONE record if found...
    if ($check !== false) {
        $row = mysql_fetch_assoc($check);

         $_SESSION['user_id'] = $row['user_id'];

         echo "Welcome back '{$row['user_name']}'!";
    } else {
        echo 'no';
    }
} else
    echo 'no';
}

关于php - 使 php 响应在页面刷新时保持可见，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10596393/