在URL中将URL作为get参数传递[关闭]

   http://example.com/geturl.php?url=http://example.org/index.php?parafile=1698%3A1562%3A0%3A0&para_action=print_ticket&parafile=dance://here.kodas/print&token=3ec2b0d3e6e0ca152bc024cc3f30f16c

So i want each of this parameters in a different varaible in the geturl.php file. I am using the regular get url

You can use anything like this :

$url = "http://example.com/geturl.php?url=".urlencode($urlPart);

$_server[QUERY_STRING] solved my problem