尝试使用jquery .post（）发送数据

In insert.php I have the following

   <script type="text/javascript"        src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js">
   <script type="text/javascript" src="includes/js/jquery.form.js"></script>
   <script type="text/javascript">




  $(document).ready(function() {
  $(".like").click(function() {
  var data = $(this).attr('data-id');
  $.post("data.php", {'name': data}, function(response){
  $('#dv').html(response);
  });
  });
 });

// ...

 </script>
<!-- ... -->
     <?
 if (!is_bool($result) && !is_null($result)){
 while($row = mysql_fetch_array($result))
 {

 ?>
    <? $id = $row['uniqueid'];?>
    <? echo "<br>ID: " . $row['uniqueid'];?> 
    <? echo "<br>Name: " . $row['surname']; ?>  
    <? echo "<br><a href class=like id=buton data-id='$id'> Like (" . $row['likes'] . ")               </a>"; ?> 
    <? echo "<br><br>"; ?>

<!-- ... -->

in data.php I have

  <script type="text/javascript"        src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js">
  <script type="text/javascript" src="includes/js/jquery.form.js"></script>
  <script type="text/javascript">
  alert("<? echo $_POST['name']; ?>");
  </script>

I fixed a couple of things I was told were wrong but it still does't work. The alert box is blank. Please help. Thanks.

this is completely wrong... for your data.php this will not parse javascript the way you are expecting.. it will simply take in POST data and ECHO out a result.. dont try and use client-side javascript because it will not do anything

data.php: its php just simply do

<?php
   echo $_POST['name']; 
?>

in your js if you really want a popup to show then you have to wait for the results of the POST/RESPONSE and do it

  $.post("data.php", {'name': data}, function(response){
     $('#dv').html(response);
     alert(response);
  });

remember the php script you call from a POST or ajax call will only process server side code.. PHP, perl, java, C... not client-side code like javascript