尝试使用jquery .post（）发送数据

Question

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.

Answer 1

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