Question

我有档案System.out.println(something.getClass().getName());和index.php。在view.php中，有一个表单可以处理一些插入数据到MYSQL表。 index.php的功能是获取view.php中人员插入的数据表。我的问题是“是否有任何代码（php，javascript等）在插入数据后立即获取数据？”并且如果数据已插入index.php，我们可以听到index.php中播放的声音，如通知声音。

Answer 1

下面是简单的代码，请参考

首先创建一个表

CREATE TABLE IF NOT EXISTS `messageTest` (
`id` int(50) NOT NULL AUTO_INCREMENT,
`notification` varchar(255) NOT NULL,
`status` varchar(50) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

然后创建View.php，如

 <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js" type="text/javascript" charset="utf-8"></script>
 <script type="text/javascript" charset="utf-8">

 function addmsg(type, msg){

 $('#notification_count').html(msg);

 }
 function playSound(filename){   
            document.getElementById("sound").innerHTML='<audio autoplay="autoplay"><source src="' + filename + '.mp3" type="audio/mpeg" /><source src="' + filename + '.ogg" type="audio/ogg" /><embed hidden="true" autostart="true" loop="false" src="' + filename +'.mp3" /></audio>';
        }
 function waitForMsg(){

 $.ajax({
 type: "GET",
 url: "select.php",

async: true,
cache: false,
 timeout:50000,

  success: function(data){
 if(data>0){
     playSound("mymp3");
 }
  addmsg("new", data);
  setTimeout(
  waitForMsg,
 1000
 );
},
 error: function(XMLHttpRequest, textStatus, errorThrown){
 addmsg("error", textStatus + " (" + errorThrown + ")");
 setTimeout(
 waitForMsg,
 15000);
}
});
};

 $(document).ready(function(){

 waitForMsg();

 });

 </script>
 <span id="notification_count"></span>
<a href="#" id="notificationLink" onclick = "return getNotification()">Notifications</a>
<div id="HTMLnoti" style="textalign:center"></div>

然后选择.php

   $servername = "localhost";
   $username = "root";
   $password = "";
   $dbname = "mydatabaseName";

   // Create connection

   $conn = new mysqli($servername, $username, $password, $dbname);

   // Check connection

   if ($conn->connect_error) {

       die("Connection failed: " . $conn->connect_error);

   } 

   $sql = "SELECT * from messageTest where status = 'unread'";
   $result = $conn->query($sql);
   $row = $result->fetch_assoc();
   $count = $result->num_rows;
   echo $count;
   $conn->close();

然后是index.php

   $servername = "localhost";
   $username = "root";
   $password = "";
   $dbname = "mydatabaseName";

   // Create connection

   $conn = new mysqli($servername, $username, $password, $dbname);

   // Check connection

   if ($conn->connect_error) {

       die("Connection failed: " . $conn->connect_error);

   } 

     $sql = "INSERT INTO messageTest (id, notification, status) VALUES (1, 'New notification', 'unread')";
   $result = $conn->query($sql);

   $conn->close();

可以插入Javascript / PHP检测数据吗？

1 个答案: