需要帮助检查用户是否在线

Question

我对PHP和MySQL相当陌生，所以我需要一些帮助。

     <?php 
         function Agotime($date) 
         { 
          if(empty($date)) { 
           return "No date provided"; 
          } 

          $periods   = array("second", "minute", "hour", "day", "week", "month", "year", "decade"); 
          $lengths   = array("60","60","24","7","4.35","12","10"); 

          $now    = time(); 
          $unix_date   = strtotime($date); 

           // check validity of date 
          if(empty($unix_date)) {  
           return "Unknown"; 
          } 

          // is it future date or past date 
          if($now > $unix_date) {  
           $difference  = $now - $unix_date; 
           $tense   = "ago"; 

          } else { 
           $difference  = $unix_date - $now; 
           $tense   = "from now"; 
          } 

          for($j = 0; $difference >= $lengths[$j] && $j < count($lengths)-1; $j++) { 
           $difference /= $lengths[$j]; 
          } 

          $difference = round($difference); 

          if($difference != 1) { 
           $periods[$j].= "s"; 
          } 

          return "$difference $periods[$j] {$tense}"; 
         } 

         $date = $run_user['lastloggedin']; 
         $result = Agotime($date); // 2 days ago 
         $serverjoins = $run_user['server_joins']; 
         ?> 

我有一个代码，它的工作，因为它应该的，但有一个问题，我有另一行我的数据库在线调用，如果设置为true，我希望它显示在线而不是例如Last Seen：1小时前 任何人都可以告诉我在哪里放什么？

Answer 1

我假设$ run_user包含结果从您的数据库，除了lastloggedin领域有一个在线场：

if ('true' == $run_user['online'] { 
    $result = 'Online now!'; 
} else { 
    $date = $run_user['lastloggedin']; 
    $result = Agotime($date); 
}