显示在线用户

Question

我想了解您如何显示在线用户？一旦他们登录，我希望它显示在我的会员页面上，现在我得到了多少用户。如果用户关闭浏览器或注销，我也希望超时，它不会显示任何在线用户，这是所有成员的代码...

<? 

if ($_GET[user] != "") { 

$check_query = mysql_query("SELECT * FROM users WHERE username='$lowerusername'"); 

if (mysql_num_rows($check_query) != "0") {

} else {

header("Location: /usernotfound.php");



} 

}

?>
<?
$username = $_SESSION['membersusername'];
$check_query3 = mysql_query("SELECT * FROM users WHERE username='$username'");
while ($display = mysql_fetch_array($check_query3)) {
    $bio = $display['bio'];


}
?>
<?

if ($_GET[user] != ""){

$check_query = mysql_query("SELECT * FROM users WHERE username='$lowerusername'");

  while ($display = mysql_fetch_array($check_query)) {

$twitter = $display['twitter'];
    $facebook = $display['facebook'];
$tinychat = $display['tinychat'];
$vaughnlive= $display['vaughnlive'];
$uploadbanner = $display['uploadbanner'];
$uploadbg= $display['uploadbg'];
$backgroundcolor = $display['backgroundcolor'];
$bgp = $display['bgp'];
$bgr = $display['bgr'];
$bga = $display['bga'];
$btc = $display['btc'];
$image = $display['image'];
            $status1 = $display['status']; 

    $djname2 = ":: $djname2 Profile";

$djname3 = "$djname3";

    $djname = "<strong>Name:</strong> $name";

        $age = "<strong>DOB:</strong> $age1";

        $email = "<strong>E-Mail:</strong> <a href='mailto:$email1'>$email1</a>";

        $year = date('d/m/Y');

        $totalage = $year-$age1;



}}

?>

<?

if ($_GET[user] != ""){

$check_query = mysql_query("SELECT * FROM profiles WHERE username='$lowerusername'");

  while ($display = mysql_fetch_array($check_query)) {

$twitter = $display['twitter'];
    $facebook = $display['facebook'];
$tinychat = $display['tinychat'];
$vaughnlive= $display['vaughnlive'];
$uploadbanner = $display['uploadbanner'];
$uploadbg= $display['uploadbg'];
$backgroundcolor = $display['backgroundcolor'];
$bgp = $display['bgp'];
$bgr = $display['bgr'];
$bga = $display['bga'];
$btc = $display['btc']; 
$username = $display['username'];
$image = $display['image'];

 }}
?>
<? $check_query = mysql_query("SELECT * FROM profiles WHERE username='$_SESSION[membersusername]'");

  while ($display = mysql_fetch_array($check_query)) {

$image1 = $display['image'];

        if ($image1 == "") { 



    $image1 = "profile/nophoto.gif";

    } else {

        $image1 = "profile/$username/$image1";



        }} 

    ?>
     <?

$check_query_all = mysql_query("SELECT * FROM users WHERE hiddenauth='no' AND auth='yes' ORDER BY id DESC");
while ($display = mysql_fetch_array($check_query_all)) {
$allmembers = $display['name'];
$allusername = $display['username'];
$allbio = $display['bio'];
$allage = $display['age'];
// get more stuff above if u need
$check_query_all2 = mysql_query("SELECT * FROM profiles WHERE username='$allusername'");
while ($display = mysql_fetch_array($check_query_all2)) {
$allimage = $display['image'];
$alllocation = $display['location'];

// Check Album Photo's
    $check_amount = mysql_query("SELECT * FROM users WHERE hiddenauth='no' AND auth='yes'");
    $num_djs_total = mysql_num_rows($check_amount);

// CHECK IF IMAGE OR NOT
if ($allimage == "") {
$allcheckedimage = "nopic.gif"; }

// CHECK IF IMAGE OR NOT
if ($allimage != "") {
$allcheckedimage = "$allusername/$allimage"; } 


// CHECK IF location OR NOT
if ($alllocation == "") {
$allcheckedlocation = "No Idea"; }

// CHECK IF location OR NOT
if ($alllocation != "") {
$allcheckedlocation = "$alllocation"; } 



// CHECK IF mini bio OR NOT
if ($allbio == "") {
$allcheckedbio = "Something about your show."; }

// CHECK IF mini bio OR NOT
if ($allbio != "") {
$allcheckedbio = "$allbio"; } 


$listalldjs .= "<table class='inlineTable' width='452' border='0'>
  <tr>
    <td width='20%'><img src='$domain/profile/$allcheckedimage'  width='180' height='173'/></td>
    <td width='83%'><b>Username:</b> $allusername<br /><b>Location:</b> $allcheckedlocation<br /><b>Show Description:</b><br>$allcheckedbio<div align='right'><a href='$domain/$allusername'/>View Channel</a>
</div> </td>
  </tr> 
</table>";
}}


?>

这是登录代码...

<?php session_start(); ?>
<? include "includes/dbconfig.php"; ?>
<? include "includes/loggedin_config.php"; ?>
<?
$lowerusername = strtolower($_POST[username]);
// MEMBERS LOGIN CHECK IF USER EXISTS :D

if ($_GET[action] == "login") {

$check_query = mysql_query("SELECT * FROM users WHERE username='$lowerusername' AND password='$_POST[password]' AND auth='yes'");

         if (mysql_num_rows($check_query) != "1") {

        $error = "Incorrect details! or You have not authorized your account!";

        } else {



$_SESSION[ipaddress] = $_SERVER['REMOTE_ADDR'];

$_SESSION[membersusername] = $lowerusername;

 header('Location: main.php');
 }

 }

?>

谁能帮我制作一个仅限在线的脚本而不是全部成员？如果是这样，我需要向 MySQL 表添加什么？

谢谢

Answer 1

注意

mysql_* 函数已折旧。使用 PDO 或 MySQLi

确保 MySQL 服务器和 PHP 服务器上的时区是 UTC。

将seen 列添加到您的 MySQL 用户表中。

当登录用户访问您网站上的任何页面时，将当前用户的“seen”行更新为 SQL 函数NOW()：

$username = $_SESSION['membersusername'];
if(isset($username)){
 mysql_query("UPDATE users SET seen=NOW() WHERE username='$username'");
}

如果用户只是停留在您的页面中，则向 seen 行已更新的页面发送 AJAX 请求。

在要显示在线用户的页面上，循环访问用户，检查“seen”行是否小于当前时间 - 20 秒。

如果少于 20 秒，则该用户处于离线状态。否则用户在线。 示例

$sql=mysql_query("SELECT seen FROM users");
while($rows=mysql_fetch_array($sql)){
 $seen=$rows['seen'];
 if($seen < date("Y-m-d H:i:s",strtotime('-20 seconds', time()))){
  echo "User Is offline";
 }else{
  echo "User is online";
 }
}

您可以将此 20 秒设置为更多或更少的时间。

Answer 2

你需要一个 cron。

每隔10秒，检查用户是否登录？

step1.当用户登录时，在表格字段中输入标志值“yes”。

第 2 步。每 10 秒检查用户是否已登录：您可以在用户注销或用户关闭选项卡时触发查询，并使用“否”标志更新完全相同的字段。

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