php在客户端禁用cookie时让session不失效的解决方法
cookie固然好,不过有些客户端浏览器会禁用cookie,这就会导致你所依赖cookies的程序会失效或出错,那么若真出现用户关闭cookies的情况,PHP应该如何再次使用session?方法还是有的。
1、设置php.ini的session.use_trans_sid = 1或者打开enable-trans-sid选项,让PHP自动跨页传递session id。
2、手动通过URL传值、隐藏表单传递session id。
3、用文件、数据库等形式保存session_id,在跨页过程中手动调用。
途径1举例说明:
s1.php
1
2session_start();
3$_SESSION[’var1’]="源码爱好者";
4$url="下一页";
5echo $url;
6?>
s2.php
1
2session_start();
3echo "传递的session变量var1的值为:".$_SESSION[’var1’];
4?>
运行以上代码,在客户端cookie正常的情况下,应该可以在得到结果“源码爱好者”。
若此时关闭了客户端的cookie,估计就得不到结果了,这时可设置php.ini中的session.use_trans_sid = 1或者编译时打开--enable-trans-sid选项”,此时又可重新得到结果“源码爱好者”
途径2举例说明:
s1.php
1
2session_start();
3$_SESSION[’var1’]="源码爱好者";
4$sn = session_id();
5$url="下一页";
6echo $url;
7?>
s2.php
1
2session_id($_GET[’s’]);
3session_start();
4echo "传递的session变量var1的值为:".$_SESSION[’var1’];
5?>
隐藏表单的方法基本原理同上。
途径3举例说明:login.html
mylogin1.php
01
02$name=$_POST[’name’];
03$pass=$_POST[’pass’];
04if(!$name || !$pass) {
05echo "用户名或密码为空,请重新登录";
06die();
07}
08if (!($name=="youngong" && $pass=="123") {
09echo "用户名或密码不正确,请重新登录";
10die();
11}
12//注册用户
13ob_start();
14session_start();
15$_SESSION[’user’]= $name;
16$psid=session_id();
17$fp=fopen("e:\tmp\phpsid.txt","w+";
18fwrite($fp,$psid);
19fclose($fp);
20//身份验证成功,进行相关操作
21echo "已登录
";
22echo "下一页";
23?>
mylogin2.php
01
02$fp=fopen("e:\tmp\phpsid.txt","r";
03$sid=fread($fp,1024);
04fclose($fp);
05session_id($sid);
06session_start();
07if(isset($_SESSION[’user’]) && $_SESSION[’user’]="laogong" {
08echo "已登录!";
09}
10else {
11//成功登录进行相关操作
12echo "未登录,无权访问";
13echo "请登录后浏览";
14die();
15}
16?>
请关闭cookie再测试,用户名:youngong 密码:123 这是通过文件保存session id的,文件是:e:\tmp\phpsid.txt。至于用数据库的方法,就不举例子了,与文件的操作方法类似。以上方法有一个共同点,就是在前一页取得session id,想办法传递到下一页,在下一页的session_start();之前加代码session_id(传过来的session id);希望能为您提供一些参考。

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