我试图传递一个包含完整url的链接;
http://xxx.co.uk/trackit.php?page=http://xxx.co.uk/game?p=buildings&menu
正如你在上面的链接中看到的,它首先转到一个名为trackit的脚本,进行一些跟踪,然后重定向到我想要的实际页面,但是在这种情况下,'&menu‘正在被删除,我想我知道为什么,但一直无法修复它。
$next_page = mysqli_real_escape_string($con, $_GET['page']);
header("Location: $next_page");
任何建议都是很好的。谢谢
发布于 2018-07-23 02:36:00
//before display your url
$page = urlencode('http://xxx.co.uk/game?p=buildings&menu');
$url = 'http://xxx.co.uk/trackit.php?page=' . $page;
echo $url;
// displays
// http://xxx.co.uk/trackit.php?page=http%3A%2F%2Fxxx.co.uk%2Fgame%3Fp%3Dbuildings%26menu
//when you receive it on the get
$next_page = urldecode($_GET['page']);
header("Location: $next_page");
https://stackoverflow.com/questions/51467958
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