Question

我知道我可能会以最糟糕的方式做到这一点，而且最重要的是它现在还没有工作。

所以我要做的是在每个用户上线时显示通知。我有一些阿贾克斯：

<!-- display online status notification Script -->
<script>
    $(document).ready(function(){  
        $.ajax({
            url: 'check-online-status.php',
            cache: false,
            error: function(e){
                alert(e);
            },
            success: function(response){
                // Reload the page to refresh data from database
               window.setTimeout(function(){ document.location.reload(true); }, 10000);
            }
        });   
    });
</script>

将检查check-online-status.php文件中的哪些用户在线：

<?php
//// require the db config file ////
require("../config.php");
//// Get users ////
$query5 = "SELECT online_status, last_activity FROM users ";
$stmt5 = $db->prepare($query5);
$stmt5->execute();
$recent_user = $stmt5->fetchAll();
?>

<?php foreach($recent_user as $row) { ?>
<?php
    // Check online status
    if($row['online_status'] == 'Online'){
        $online_status = 'online';
    } 

    // Get last activity date/time and compare to todays date/time
    if(!is_null($row['last_activity'])){
        $start_date = new DateTime($row['last_activity']);
        $end_date = new DateTime(date("Y-m-d h:i:sa"));
        $interval = $start_date->diff($end_date);
        $last_activity = "Last Active: " . $interval->y . " years, " . $interval->m." months,  ".$interval->d." days, ".$interval->h." hours, ".$interval->i." minutes "."and ".$interval->s." seconds "."ago";

        if($interval->s == 0){
            $last_activity = "Last Active: Now";
        }
    }
    else {
        $last_activity = 'None';
    }

?>  
<?php if($online_status == 'online' && $last_activity == 'Last Active: Now' ){ ?>
    <!-- Launch a SmallBox on form load -->
    <script type="text/javascript"> 
        $(document).ready(function(){
            var username = <?php echo ''.$row['username'].''.'is online'; ?>;
            $.bigBox({
                title: "Online",
                content: username,
                width: 250,
                color: "#2ECC71",
                timeout: 5000,
                delay: 0.5,
            });


        }); 
    </script>
 <?php } ?>

我正在使用名为Metro Notification的脚本来显示通知框。

不确定我做错了什么。当我用firebug检查时，我可以通过JS（位于check-online-status.php文件中）看到信息看起来正确：

<!-- Launch a SmallBox on form load -->
    <script type="text/javascript"> 
        $(document).ready(function(){
            var username = user1 is online;
            $.bigBox({
                title: "Online",
                content: username,
                width: 250,
                color: "#2ECC71",
                timeout: 5000,
                delay: 0.5,
            });


        }); 
    </script>


        <!-- Launch a SmallBox on form load -->
    <script type="text/javascript"> 
        $(document).ready(function(){
            var username = user2 is online;
            $.bigBox({
                title: "Online",
                content: username,
                width: 250,
                color: "#2ECC71",
                timeout: 5000,
                delay: 0.5,
            });


        }); 
    </script>

如果2个用户在线，这是我上面的代码。

用户上线时显示通知

0 个答案: