如何获取PHP以显示从浏览器收到的标头？
以下代码行可用于显示通过浏览器接收的PHP代码的标头-
<?php print_r($_SERVER[URL]) ?>
要么
示例
$headers = getallheaders();
foreach($headers as $key=>$val){
echo $key . ': ' . $val . '<br>';
}输出结果
这将产生以下输出-
Host: www.websitename.com Content-Length: 180 Cache-Control: max-age=0 Origin: http://www.websitename.com Upgrade-Insecure-Requests: 1 Content-Type: application/x-www-form-urlencoded User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/79.0.3945.79 Safari/537.36 Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9 Referer: http://www.writephponline.com/ Accept-Encoding: gzip, deflate Accept-Language: en-GB,en-US;q=0.9,en;q=0.8 Cookie: _ga=GA1.2.118418268.1575911248; _gid=GA1.2.1794052867.1576852059