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Как это правильно настроить?
<?php
$db_user = "root";
$db_pass = "outoftouch";
$db_name = "l2jdb";
$db_serv = "213.149.129.175";
$res = mysql_connect ( $db_serv, $db_user, $db_pass ) or die ("Coudn't connect to [$db_serv]");
$resdb = mysql_select_db ( "$db_name",$res );
return $res;
?>
Имя БД: game
login: root
pasw: seraqer
ip: 127.0.0.1
Вот он связан с таким:
<?php include "config/stats.php" ?>
<?php
mysql_connect($db_serv,$db_user,$db_pass) or die (mysql_error());
mysql_select_db($db_name) or die(mysql_error());
//Online Players
$query = ("SELECT * FROM characters WHERE online=1 and accesslevel>=0");
$result = mysql_query($query) or die(mysql_error());
$num2=mysql_numrows($result);
$sum=$num2;
//Total Accounts
$query = ("SELECT * FROM accounts WHERE access_level>=0");
$result = mysql_query($query) or die(mysql_error());
$acc2=mysql_numrows($result);
//Total Characters
$query = ("SELECT * FROM characters WHERE accesslevel>=0");
$result = mysql_query($query) or die(mysql_error());
$char2=mysql_numrows($result);
//Alliance
$query = ("SELECT * FROM clan_data WHERE ally_id!=\"\"");
$result = mysql_query($query) or die(mysql_error());
$alliance2=mysql_numrows($result);
//Clan
$get_config_data="SELECT clan_id FROM clan_data";
$query = ("SELECT clan_id FROM clan_data WHERE clan_id>0");
$result = mysql_query($query) or die(mysql_error());
$clan2=mysql_numrows($result);
mysql_close();
mysql_connect($db_serv,$db_user,$db_pass) or die (mysql_error());
mysql_select_db($db_name) or die(mysql_error());
?>
Total Accounts: <?php print ("$acc2"); ?><br>
Total Characters: <?php print ("$char2"); ?><br>
Total Aliances: <?php print ("$alliance2"); ?><br>
Total Clans: <?php print ("$clan2"); ?><br>
Добавлено (03.01.2013, 13:47)
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UP!!!
Добавлено (03.01.2013, 15:44)
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решил проблему вырезав эту фигню в index.php
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