通过ajax在网站上提醒用户

Question

我在尝试制作通知系统，通知用户将其分配给故障单或将新故障单添加到数据库时。

我已经拥有的系统本身的工作原理除了它只将通知发送给接收ajax请求的第一个用户。有什么办法可以让每个想要收到通知的人都能收到通知？

我的代码： 的javascript：

 function checkUpdates() 
      { 
       $.ajax({ 
        type: "POST", 
        url: 'ajax/checkDB.php', // a webservice or other URL that queries the database 
        data: {}, 
        contentType: "application/json; charset=utf-8", 
        dataType: "json", 
        success: function(data) { 
         console.log('Connected and executed PHP page... '+data); 
         if (data.hasChanged == "true") { 
          playSound('img/notif2'); 
          notifyAdmin(); 
          console.log('Updated Tickets Page...'); 
          $("#contents").load("dynamic/tickets.php"); 
          $("#contents").fadeTo("fast", 1); 
         } 

         if (data.newAssigned == "true") { 
          playSound('img/notif2'); 
          notifyUser(); 
          console.log('Updated Tickets Page...'); 
          $("#contents").load("dynamic/tickets.php"); 
          $("#contents").fadeTo("fast", 1); 
         } 
        } 
       }); 
      } 
     $(document).ready(function() { 
      setInterval("checkUpdates()", 3000); // Calls the function every 3 seconds 
     }); 

我的PHP脚本（checkDB.php）：

<?php 
include("../static/config.php"); 
session_start(); 

header("Content-Type: text/json"); 

     $result = mysqli_query($con,"SELECT notify FROM tickets WHERE notify='0'"); 
     $row_cnt = mysqli_num_rows($result); 

     if($row_cnt > 0) { 
      $hasChanged = 'true'; 
      mysqli_query($con,"UPDATE tickets SET notify='1' WHERE notify='0'"); 
     } else { 
      $hasChanged = 'false'; 
     } 

     $result2 = mysqli_query($con,"SELECT notify,Assigned FROM tickets WHERE Completeness='1' AND notify='1'"); 
     $row_cnt2 = mysqli_num_rows($result2); 

     if($row_cnt2 > 0) { 
      while($row2 = mysqli_fetch_array($result2)) 
       { 
        if(strcmp($_SESSION['Name'],$row2['Assigned']) == 0) { 
         $newAssigned = 'true'; 
        } else { 
         $newAssigned = 'false'; 
        } 
       } 
      mysqli_query($con,"UPDATE tickets SET notify='2' WHERE Completeness='1' AND notify='1'"); 
     } else { 
      $newAssigned = 'false'; 
     } 

     echo json_encode(array('newAssigned' => $newAssigned, 'hasChanged' => $hasChanged)); 

?> 

这里是我的确切意图的破败： 用户登录到系统中，并给予行政权利。 用户加载票据页面，其中列出了所有票据以及我给你的这个javascript JavaScript每3秒加载一次checkDB.php，基本上运行一个针对数据库的'notify'值为0的检查，并且如果在至少存在1个，在数据库中更新为'1'，并返回为正确的ajax调用 - 发送通知。

再次，这适用于加载到页面的第一个用户，但在此之后，显然在数据库中不再有'notify = 0'，因为它们已经被更新，所以它们不会显示通知。

有人要求我分享我的数据库结构。 门票：

UniqueID - Unique ID per ticket (always unique) 
Requester - Whoever submitted the ticket. 
Problem - Description of the given issue. 
Assigned - User who is assigned to the ticket. 
Completeness - The level of completeness (0-4) 
Times - Times listed for ticket start, assigned, etc 
notified - Used in old ticket system (I plan to get rid of it) 
Urgency - Selected by requester, how urgent the ticket is 
Location - Location in our warehouse assistance is needed. 
FollowUp - currently not in use, will be used to submit follow ups. 
isProject - (0-1) is project or not 
additionalTechs - Lists additional users on a ticket (Not important for question) 
notify - (0-2) at 0, ticket is new and notif should be sent to admins which should set this to 1, when it is 1 it should send a notif to users that are attached to the ticket. 

科技股数据库：

UniqueID 
Username 
Password 
Name 
Level 
Disabled 
LastTimeSeen 
needsNotify (Used in old system, plan to remove) 

我真的被困在这里。

Answer 1

你可以使用websockets或socket.io。 那么多个客户端的JS将与插座连接会通知所有连接的人们了解他们票

Answer 2

你可以改变你的更新语句是这样的：

UPDATE tickets SET notify='1' WHERE notify='0' and Assigned = ''//some user id passed