使 PHP 响应在页面刷新时保持可见

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';  
}

$(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;
});
});

问题是JS只是客户端脚本语言 - 它仅在客户端浏览器上处理。

如果您想使用 AJAX 登录，好的，但您必须将登录用户的值存储到会话(或 cookie(中。然后，在每个页面加载时，您将检查该会话或cookie是否已设置和存在，如果是，您将在登录后写下与jQuery相对应的HTML...

换句话说，这是你将要做的：

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. 您检查该用户是否存在，如果存在，则保存$_SESSION['username'](例如(
4. 在 AJAX 中，您可以填补该$('#login_reply')
5. 用户点击您网站中的某些链接
6. 检查您是否在$_SESSION['username']中设置了值(或索引(
7. 如果是，通过 PHP 您将填写 #login_replydiv，如果不是，您将看到登录表单...

希望这有帮助...

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，因此应将正常的POST计算在内...

编辑2：我还想指出一些一般的编程错误...

这是您的代码，添加了我的注释：

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(仍在使用 mysql_*，对此表示抱歉(：

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';
}

最简单的解决方案是使用 cookie：

在PHP端，你声明并删除cookie：

<?php
if (!isset($_COOKIE['new_one']) ) {
  setcookie('new_one', "ole", 0, "/");
  echo "logged in";
}
else {
  setcookie('new_one', null);
  echo "logged out";
}
?>

在jQuery客户端：

$(document).ready(function() {
  if ($.cookie("new_one") === "ole") {
    $('#login_reply').html("Show msg only when the server sets the cookie.");
  }
});

您需要将登录回复数据存储在服务器上的某个位置或会话中; JavaScript 是一种客户端语言，因此在刷新之间不会保留任何实现的值。

我建议你存储结果，然后在用php生成页面时检查它是否存在;如果有，填写数据......等等。