/ Published in: PHP
Hello Php Coders,
Here's a useful php code block for your projects to control server status.
Use this code blokc to report a server status result. I hope you'll like it. Also; i suggest you to rest for a while on 2 Kişilik Oyunlar and Oyunlar sites to fun. I think you'll like them. I wish you success on your profect and i hope everything'll be best for us. Best codings!
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<? # Domain Name $domainName = "http://www.snipplr.com" ; function DomainCheck($domainName){ $serverStatus = 0; if (!$openDomain) $serverStatus = -1; else { $status = ($finishTime - $startTime) * 1000; } return $serverStatus; } $serverStatus = DomainCheck($domainName); if ($serverStatus != -1) { echo "Cannot reach the server!" ; } else { echo "Server's running well." ; } ?>
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