ziyaretciler hangi sayfada oldugunu gösteren kod lazım realistteki gibi. yardımcı olacak olan varmı
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CREATE TABLE `online` (
`id` bigint(20) NOT lisanssız auto_increment,
`timestamp` int(15) NOT lisanssız default '0',
`ip` varchar(40) NOT lisanssız default '',
`file` varchar(100) NOT lisanssız default '',
PRIMARY KEY (`id`),
KEY `ip` (`ip`),
KEY `file` (`file`),
KEY `timestamp` (`timestamp`)
) TYPE=MyISAM;
<?
$db_host = "localhost";
$db_user = "kullanici";
$db_pass = "123456";
$db_name = "online";
$connect = @mysql_connect($db_host,$db_user,$db_pass);
$db = mysql_select_db($db_name,$connect);
if (!$connect) {
echo ("noconnection");
exit();
}
?>
<?php
include('baglanti.php');
//Fetch Time
$timestamp = time();
$timeout = $timestamp - 180;
//Insert User
$insert = mysql_query("INSERT INTO online (timestamp, ip, file) VALUES('$timestamp','".$_SERVER['REMOTE_ADDR']."','".$_SERVER[PHP_SELF].'?'.$_SERVER[QUERY_STRING]."')") or die("Error in who's online insert query!");
//Delete Users
$delete = mysql_query("DELETE FROM online WHERE timestamp<$timeout") or die("Error in who's online delete query!");
//Fetch Users Online
$result = mysql_query("SELECT DISTINCT ip FROM online") or die("Error in who's online result query!");
$users = mysql_num_rows($result);
//Show Who's Online
if($users == 1) {
print("Online $users\n");
} else {
print("Online $users\n");
}
?>
<?php
include('baglanti.php');
$sorgu = mysql_query("SELECT file,timestamp, ip FROM online ORDER BY id DESC" );
while ($kayit = mysql_fetch_array($sorgu, MYSQL_ASSOC))
{
$file = $kayit["file"];
$ip = $kayit["ip"];
$saat = $kayit["timestamp"];
$date = date("d.m.Y - H:i:s", $saat);
echo "<div><table border='0' width='100%' height='19' style='border-collapse: collapse'><tr><td height='19' width='120'><font size='2'>".$date." -</font></td><td height='19' width='80'><font size='2'>".$ip."</font></td><td height='19' width='600'><a href ='".$file."' target='_blank'><font size='2'>".$file."</font></td></tr></table></div>";
}
?>