Php 迭代mysql结果以进行比较

这是检查服务器是否打开/关闭的脚本

$ip = "xxxxxxxx";
$port = "xxx";
$fp = @fsockopen("udp://$ip", $port, $errno, $errstr, 1);
socket_set_timeout($fp, 000002); 
if (!$fp) 
{ 
  echo "OFFLINE"; 
}
else 
{ 
  echo "ONLINE"; 
} 
@fclose($fp); 

mysql表

id    |  ip    | port      |  online_status | 
1     |  ip1   | port1     |                |
2     |  ip2   | port2     |                |
3     |  ip3   | port3     |                |
4     |  ip4   | port4     |                |
5     |  ip5   | port5     |                |

我需要检查每个服务器是否同时打开/关闭，并在表中写入。
我该怎么做

$servers = $db->getArray("SELECT * FROM servers");

---

尝试在阵列中循环并尝试连接到每台服务器。这将告诉您是否可以连接到服务器。然后，更新表以包含状态

/* Loop through the servers */
foreach($servers as $server)
{
   $id = $server['id'];
   $ip = $server['ip'];
   $port = $server['port'];
   /* Try to connect to this particular server */
   $fp = @fsocketopen("udp://$ip", $port, $errno, $errstr, 1);

   socket_set_timeout($fp, 000002);

   $status = '';
   /* Determine the status of the server */
   if(!$fp)
   {
      $status = 'OFFLINE';
   }
   else
   {
      $status = 'ONLINE';
   }
   /* Close our connection */
   @fclose($fp);

   /* Update the servers table to show the status of the server */
   $db->Update("UPDATE servers SET online_status='$status' WHERE id='$id'");
}