【发布时间】：2012-12-02 11:08:19
【问题描述】：
请帮助我一次只验证一个会话，请参阅下面的脚本，该脚本当前允许相同的用户名登录任意数量的会话。
我不确定何时何地验证会话，请帮助我只添加可以验证用户名会话的那几行。
<?php // accesscontrol.php
include_once 'common.php';
include_once 'db.php';
session_start();
$uid = isset($_POST['uid']) ? $_POST['uid'] : $_SESSION['uid'];
$pwd = isset($_POST['pwd']) ? $_POST['pwd'] : $_SESSION['pwd'];
if(!isset($uid)) {
?>
<!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Login</title>
<meta http-equiv="Content-Type"
content="text/html; charset=iso-8859-1" />
<head>
<style type="text/css">
<!--
.style1 {
font-size: 16px;
font-family: Verdana, Arial, Helvetica, sans-serif;
}
.style3 {
font-size: 12px;
font-family: Verdana, Arial, Helvetica, sans-serif;
}
body {
background-color: #D7F0FF;
margin-left: 0px;
margin-top: 0px;
margin-right: 0px;
margin-bottom: 0px;
}
-->
</style>
</head>
<body>
<h1 class="style1"> <br><br>Amogh Site - Login Required </h1>
<span class="style3"><br>
You <strong>must login to access this area </strong>of the site. <br>
<br>
If you are not a registered user, please contact your Admin
to sign up for instant access!</span>
<p><form method="post" action="<?=$_SERVER['PHP_SELF']?>">
<span class="style3">User ID:
<input type="text" name="uid" size="12" />
<br>
<br />
Password:</span>
<input type="password" name="pwd" SIZE="12" />
<br>
<br />
<input type="submit" value="Login" />
</form></p>
</body>
</html>
<?php
exit;
}
$_SESSION['uid'] = $uid;
$_SESSION['pwd'] = $pwd;
dbConnect("hitek_svga3");
$sql = "SELECT * FROM user WHERE
userid = '$uid' AND password = '$pwd'";
$result = mysql_query($sql);
if (!$result) {
error('A database error occurred while checking your '.
'login details.\\nIf this error persists, please '.
'contact you@example.com.');
}
if (mysql_num_rows($result) == 0) {
unset($_SESSION['uid']);
unset($_SESSION['pwd']);
?>
<!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title> Access Denied </title>
<meta http-equiv="Content-Type"
content="text/html; charset=iso-8859-1" />
<style type="text/css">
<!--
.style1 {
font-size: 16px;
font-family: Verdana, Arial, Helvetica, sans-serif;
}
.style3 {
font-size: 12px;
font-family: Verdana, Arial, Helvetica, sans-serif;
}
-->
</style>
</head>
<body>
<br/>
<br/>
<h1 class="style1"> Access Denied </h1>
<p class="style3">Your user ID or password is incorrect, or you are not a
registered user on this site. To try logging in again, click
<a href="<?=$_SERVER['PHP_SELF']?>">here</a>. To access, please contact our Admin !</a>.</p>
</body>
</html>
<?php
exit;
}
$username = mysql_result($result,0,'fullname');
$_SESSION['user'] = mysql_result($result,0,'userid');
$_SESSION['email'] = mysql_result($result,0,'email');
$_SESSION['notes'] = mysql_result($result,0,'notes');
?>
【问题讨论】：
-
您的脚本似乎容易受到SQL injections 的攻击。看看Best way to prevent SQL injection?
-
由于 HTTP 是无状态的，因此很难确定用户的会话是否已过期，从而确定新的登录尝试是新会话还是现有会话的副本。
-
@eggyal 然后可以简单地删除任何以前的会话并每次创建一个新会话。
-
@ose：是的。也许我误解了这个问题，因为我的阅读意图是拒绝进一步的登录尝试，而不是退出现有会话。
-
顺便说一句，作为一般建议，在发布代码时删除公司名称等可识别信息是明智的。您永远不知道谁会在以后发现此代码，并推测您很容易受到特定风格的 SQL 注入攻击并侵入您的系统...