获取 http 状态响应码的有效方法

Question

Is there a quicker and more efficient way to check if a website shows status codes 200 etc?

Basically I want to check specific pages whether they're online of offline. Some of the pages I check go down sporadically with 502 errors, so I need a way to also make sure those are displayed as "offline" thus using the header response check.

when using the below function on say 5-10 pages, it slows the page loading time dramatically.

<?php
function getHttpResponseCode( string $url ) //: int
{
$headers = get_headers( $url );
return substr( $headers[ 0 ], 9, 3 );
}

if (getHttpResponseCode('https://www.google.com') !='200') {
echo 'offline';
} else {
echo 'online';
}
?>

**UPDATED CODE (not working atm) **

 <?php
function checkDomain($host) {

if($socket =@ fsockopen($host, 80, $errno, $errstr, 30)) {
  echo 'online!';
  fclose($socket);
} else {
  echo 'offline.';
}
}
echo checkDomain('https://www.google.com');
?>

Answer 1

The speed should inprove considerably using this:

$host = 'google.com';
if($socket =@ fsockopen($host, 80, $errno, $errstr, 30)) {
  echo 'online!';
  fclose($socket);
} else {
  echo 'offline.';
}