Question

如果我有一个带有unix time（）标记的列“lastaccess”，我该如何过滤掉在线用户。然后我通常有这个功能：

function showStatus($userid) {
    $timeoutIdle = 80;
    $timeoutOffline = 150;
    $row = mysql_query("select last_access from users where id='$userid'");
    $read = mysql_fetch_array($row);
    $last_access = $read["last_access"];
    $thetime = time();
    if ($thetime - $last_access > $timeoutOffline) {
        echo "";
    }
    else if ($thetime - $last_access > $timeoutIdle) {
        echo "<span class='statusIdle'>Idle</span>";
    } else {
        echo "<span class='statusOnline'>Online</span>";
    }
}

如何进行查询，如果您在线，它会显示在哪里？

Answer 1

$sql = "SELECT
          CASE
            WHEN last_access < UNIX_TIMESTAMP()-150 THEN 'offline'
            WHEN last_access < UNIX_TIMESTAMP()-80 THEN 'idle'
            ELSE 'online'
          END AS online_status
        FROM users
        WHERE id = $userid";

Answer 2

MySQL有一些与Date/Times相关的功能。查看NOW()和UNIX_TIMESTAMP()。您可以将此添加到您的查询where子句SELECT仅限在线用户。

SELECT * FROM users WHERE NOW() - last_access < 80;

或使用CASE语句确定给定用户的状态字符串

SELECT CASE
WHEN (NOW() - last_access > 150) THEN 'offline'
WHEN (NOW() - last_access > 80) THEN 'idle'
ELSE 'online' END AS status
FROM users WHERE id = $user_id;

Answer 3

我想说一个更简单的方法是，当用户登录时更新该用户的状态字段：

$update = mysql_query("UPDATE table SET status='online' WHERE id='$userid'") 
or die(mysql_error());

然后在您要查看他们是否在线的页面上添加以下内容：

$status = mysql_query("SELECT online FROM table
 WHERE id='$searched_user'") or die(mysql_error());  
$row = mysql_fetch_array( $status );

if($row['status'] == "online"){
echo "Online";
}
else{
echo "Offline";
}

然后当他们点击退出或cookie /会话已过期时：

$update = mysql_query("UPDATE table SET status='offline' WHERE id='$userid'") 
or die(mysql_error());

PHP：显示每个用户是否在线

3 个答案: