I wrote a script like this
if ($oyun == $uo) {
$acik = fsockopen("212.15.232.14",5631);
if ($acik) {
$oyuncular = "oyuncular";
$kayitli = "kayitli";
echo "Server: Open <br>";
echo "<a href=\"oyun.php?oyun=$uo&durum=$oyuncular\" target=\"_blank\">Oyunda: $file[0] </a><br>";
echo "<a href=\"oyun.php?oyun=$uo&durum=$kayitli\" target=\"_blank\">Kayitli: $row </a><br>";
}
else {
echo "Server: Close <br><br>";
echo "Oyunda: 0 <br><br>";
echo "Kayitli: $row <br><br>";
}
}
else if ($oyun == $sc) {
$acik = fsockopen("212.15.232.15",6112);
if ($acik) {
$acikoyunlar = "acikoyunlar";
$oyuncular = "oyuncular";
echo "Sunucu: Acik <br>";
echo "<a href=\"oyun.php?oyun=$sc&durum=$oyuncular\" target=\"_blank\">Oyunda: $sayi </a><br>";
echo "<a href=\"oyun.php?oyun=$sc&durum=$acikoyunlar\" target=\"_blank\">Acik Oyun: $acikoyun </a><br>";
}
else {
echo "Sunucu: Kapali <br><br>";
echo "Oyunda: 0 <br><br>";
echo "Acik Oyun: 0";
}
}
but when 5631 port is closed its giving an error like this
Warning: fsockopen() [function.fsockopen]: unable to connect to 212.15.232.14:5631 (A non-blocking socket operation could not be completed immediately. ) in C:\Program Files\Apache Group\Apache2\htdocs\syl\functions.php on
I dont want this error to be shown
how can i prevent this, does anyone have any idea about it?
Thanks
