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php在客户端禁用cookie时让session不失效的解决方法

Jun 13, 2016 am 09:35 AM
cookie php session No exist Invalid good client method have of Disable solve

   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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