Question

我的在线用户列表存在问题。代码工作正常，所有在线用户都显示在屏幕上，但是当我点击刷新时，再次显示同一用户的电子邮件，当我第二次点击刷新时，用户的电子邮件显示三次，依此类推。 / p>

这是我的代码：

<?php

require_once("db.php");

db_connect();

session_start();


$player_timeout = time() - 5 * 60;

$time = time();

if (isset($_SESSION['email'])) {

$login=mysql_query("insert into activePlayer(player_email,time_visited,status)   values('".$_SESSION['email']."','".$time."', 'true')");

}

else

{echo "You are not logged in";}

$tmout = mysql_query("DELETE FROM activePlayer WHERE time_visited < ".$player_timeout);

$online_member = mysql_query("SELECT player_email FROM activePlayer");

$row=mysql_num_rows($online_member);

$member_row=mysql_fetch_array($online_member);

echo "Welcome &nbsp; '".$_SESSION['email']."'";

?>

<body>

<select > <?php  

 if ($row<1)
      {
        echo "&nbsp;";
      }
      else
      {?> <p><p>Online Players:<option><?php echo $member_row['player_email'];?>  
     </option>}
      <?php for ($i=1;$i<$row;$i++)

{
      $member_row=mysql_fetch_array($online_member);?>
      <p><p>Online Players:<option><?php echo $member_row['player_email']; }}?>   
 </option></select>
</body>

请问我该如何解决这个问题

Answer 1

您需要做的是重定向：

if (isset($_SESSION['email'])) {
  mysql_query("INSERT INTO activePlayer (player_email,time_visited,status)
               VALUES ('".$_SESSION['email']."','".$time."', 'true')");

  unset($_SESSION['email']);
  header("Location: otherpage.php"); // or it can be the same page
}

Answer 2

每次刷新时，如果用户已登录，则在数据库中插入一行。您必须检查用户是否已存在于数据库中并更新其记录。如果他没有记录，那么就像你一样创建一个新的。

$hasRow = mysql_query("SELECT * FROM activePlayer WHERE player_email='".$_SESSION['email']."' LIMIT 1");
if(mysql_num_rows($hasRow) > 0) {
    $login = mysql_query("UPDATE activePlayer SET visited=".time()." WHERE player_email='".$_SESSION['email']."'");
} else {
    $login=mysql_query("insert into activePlayer(player_email,time_visited,status)   values('".$_SESSION['email']."','".$time."', 'true')");
}

刷新在线用户列表中的问题

2 个答案: