使页面刷新时php响应保持可见

Question

它在我的登录脚本中正常工作。我在我的div（id = login_reply）中获得正确的响应，会话开始。但是，无论何时刷新页面，login_reply都消失了。我如何能够保持login_reply？谢谢！

这里是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'; 
} 

这里是jQuery的

$(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; 
}); 
}); 

Answer 1

问题是，JS只是客户端脚本语言 - 它只处理客户端浏览器。

如果您想使用AJAX登录，没关系，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您将检查该会话或cookie是否设置和存在，如果他们这样做，您将登录后记下与jQuery相对应的HTML ...

换句话说，这里是什么你会做：

1. 页面加载 - 用户
2. 用户没有登录填写登录凭据并点击“登录”
3. 您检查用户是否存在，如果他这样做，为您节省$_SESSION['username']（用于例如）
4. 在AJAX中填写$('#login_reply')
5. 用户点击某个环节，您的网站
6. 内您检查是否有一个值（或索引设置）在$_SESSION['username']
7. 如果它是由PHP你将填补#login_reply格，如果是不，你会显示登录表单...

希望这有助于...

EDIT1：你也应该以一种没有JS工作实现登录功能（禁用）在客户端上owser如此正常POST应该被计数...

EDIT2：我也想指出一些常见的编程错误...

这里是你的代码添加了我的意见：

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'; 
} 

这里是我重写代码（仍在使用mysql_ *，感到抱歉）：

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'; 
} 

Answer 2

你需要在一些地方保存在服务器上或者是在您的会话登录回复数据; javascript是一种客户端语言，因此任何取得的值都不会在刷新之间保持不变。

我会建议你存储结果，然后检查它是否存在，而用php生成页面;如果是这样，填写数据..等

Answer 3

最简单的解决方法是使用Cookies：

在PHP方面您声明并删除Cookie：

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

而jQuery的客户端：

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