在页面离开时销毁 PHP 会话 [英] Destroy PHP session on page leaving

问题描述

当用户离开特定页面时，我需要销毁会话.我在页面末尾使用 session_destroy() 但它对我来说不可行，因为我的页面有分页.我的页面是:abc.php?page=1 或 abc.php?page=2 或 abc.php?page=3.

I need to destroy a session when user leave from a particular page. I use session_destroy() on the end of the page but its not feasible for me because my page has pagination. My page is: abc.php?page=1 or abc.php?page=2 or abc.php?page=3.

因此，当用户从 abc.php 页面离开时，我需要销毁会话.如何在不使用 cookie 的情况下做到这一点?

So, I need to destroy a session when a user leaves from abc.php page. How can I do it without using a cookie?

推荐答案

在用户离开页面时做某事是错误的方法，因为你不知道用户是否会导航到一个完全不同的页面(比如联系.php 为了论证)，否则他/她将直接转到 abc.php 的下一页，正如 Borealid 指出的那样，没有 JS 就无法做到这一点.相反，您可以简单地添加一个检查并查看用户是否来自 abc.php:

Doing something when the user navigates away from a page is the wrong approach because you don't know if the user will navigate to a whole different page (say contact.php for the sake of the argument) or he/she will just go to the next page of abc.php and, as Borealid pointed out, you can't do it without JS. Instead, you could simply add a check and see if the user comes from abc.php:

首先，在您的 abc.php 文件中，在 $_SESSION 数组中设置一个唯一变量，该变量将作为用户已在此页面上的标记:

First, in your abc.php file set a unique variable in the $_SESSION array which will act as a mark that the user has been on this page:

$_SESSION['previous'] = basename($_SERVER['PHP_SELF']);

然后，在所有页面上添加这个，在任何输出之前检查用户是否来自 abc.php:

Then, add this on all pages, before any output to check if the user is coming from abc.php:

if (isset($_SESSION['previous'])) {
   if (basename($_SERVER['PHP_SELF']) != $_SESSION['previous']) {
        session_destroy();
        ### or alternatively, you can use this for specific variables:
        ### unset($_SESSION['varname']);
   }
}

这样，只有当用户来自 abc.php 并且当前页面是另一个页面时，您才会销毁会话(或特定变量).

This way you will destroy the session (or specific variables) only if the user is coming from abc.php and the current page is a different one.

我希望我能够清楚地解释这一点.

I hope I was able to clearly explain this.

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