Question

我正在构建一个通知系统，我只需要最后一个部分来使其完美运行。

当ajax首次加载时，它使用php变量对服务器端脚本查找新内容，如果它找到新内容，它会使用新的notification_id并更新var以获取下一组结果，如果它没有t在url notification_id周围第二次找到任何内容变为null并将空值插入div。

我如何阻止这种情况发生？

<script type="text/javascript">
var notification_id="<?php echo $notification['notification_id'] ;?>";
function loadIt() {

$.ajax({
type: "GET",
url: "viewajax.php?notification_id="+notification_id,   
global: true,
dataType:"json",
cache: false,
success: function(data){
if(notification_id < data.notification_id){
 $("#notif_actual_text-"+data.notification_id).prepend('<div class="notif_ui"><div class="notif_text"><div  id="notif_actual_text-'+data['notification_id']+'" class="notif_actual_text"><img border=\"1\" src=\"userimages/cropped'+data['notification_triggeredby']+'.jpg\" onerror=this.src=\"userimages/no_profile_img.jpeg\" width=\"40\" height=\"40\" ><br /><a href="'+data['notification_id']+'">'+data['notification_content']+' </a><br />'+data['notification_time']+'<br /></div></div></div></div>');
 i = parseInt($("#mes").text()); $("#mes").text((i+data.num)); 
    }
notification_id = data.notification_id; 
}
});
}
setInterval(loadIt, 10000);              
 </script>

Answer 1

试试这个（如果上面的代码来自viewajax.php）：

// Simply use isset() to check if $_GET['notification_id'] is set or not
var notification_id="<?php echo isset($_GET['notification_id']) ? $_GET['notification_id'] : $notification['notification_id']; ?>";

或者如果您要将其发送到另一个文件：

<?php
  // Let's just say that this is viewajax.php
  $notification_id = $_GET['notification_id'];
  echo json_encode(array("notification_id" => $notification_id));
?>

Ajax - 如果没有新数据，则获取Null值

1 个答案: