PHP如何调用动态URL
RT
$id = trim($_GET[‘Url’],'”‘);
if($id””){
$url = ‘http://www.hello-jobs.com/newsletter/’.$id.’.html1本文来#源gaodai$ma#com搞$代*码*网
搞代gaodaima码
‘;
include_once($url);
//echo $url;
测试URL是正确的,是不是我的PHP配置问题?
——解决方案——————–
; Whether to allow include/require to open URLs (like http:// or ftp://) as files.
; http://php.net/allow-url-include
allow_url_include = on
如果包含的文件是 url 则需要打开 allow_url_include 开关