Private Top List Code Error

[StreetNet]

Hello all RageZone

Why don't get Ip status look the code.

please say me why don't print status?

<?php $result = mysql_query("SELECT * FROM _Server");

echo "<table border='1'>
<tr>
<th>ServerName</th>
<th>Players</th>
<th>Level Cap</th>
<th>Experience</th>
<th>Status</th>
</tr>";

while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['ServerName'] . "</td>";
echo "<td>" . $row['Level_Cap'] . "</td>";
echo "<td>" . $row['Level_Cap'] . "</td>";
echo "<td>" . $row['Exp_Rate'] . "</td>";
echo "<td>".$ServerIP=$row['$Server_IP'];
$ServerPort=$row['$Server_Port'];
$IP = "$ServerIP";
$Port = "$ServerPort";
$status = checkState($IP,$Port);
if ($status) echo '<span style="color: green;font-weight:bold;">Online</span>';
else echo '<span style="color: red;font-weight:bold;">Offline</span>';
"</td>";
echo "</tr>";
}
echo "</table>";
?>

and only print



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Thank's

[MentaL]

[serend]

this is my code i use if it helps

function GetServerStatus($site, $port)
{
$status = array("<span style='color:red'>OFFLINE</span><b>", "<b><span style='color:green'>ONLINE</span>");
$fp = @fsockopen($site, $port, $errno, $errstr, 2);
if (!$fp) {
    return $status[0];
} else
  { return $status[1];}
}
$site = "ggg.com";
$port = "15779";
print "Server status: ";
print GetServerStatus($site, $port);