I have a URL like this;
http://www.mydomain.co.uk/blist.php?prodCodes=NC023-NC022-NC024-NCB33&customerID=NHFGR
Which i grab using HTTP Referrer. The trouble is i only need the page name ie blist.php from the link, not the entire URL as is default using:
$_SERVER['HTTP_REFERER']
Can anyone give me an idea on how to grab that part of the URL?
try with
parse_url($_SERVER['HTTP_REFERER'],PHP_URL_PATH);
Note: this variable doesn't exist in mobile devices.
Try this one:
basename($_SERVER['HTTP_REFERER']);
basename(parse_url($_SERVER['HTTP_REFERER'],PHP_URL_PATH)
这将只为您提供文件名而不是任何子目录或查询字符串
/[\w-]+.php/
您可以使用正则表达式仅获取文件的名称。
I think you are looking for $_SERVER['REQUEST_URI'] which just gives you the requested page.
You can review all of the $_SERVER variables here:
This seems to work, but there might be other elegant url functions .
$url = 'http://www.mydomain.co.uk/blist.php?prodCodes=NC023-NC022-NC024-NCB33&customerID=NHFGR';
preg_match('/\/[a-z0-9]+.php/', $url, $match);
$page = array_shift($match);
echo $page;
check out
http://ca.php.net/manual/en/function.parse-url.php
focus in on the path ($_SERVER['REQUEST_URL']) portion, then do something like substr($path,strrpos($path,"/")+1);
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