Question

I have a label and a dropdown menu where the label is dynamically changed when the dropdown menu changes. So I used ajax to solve this task but how do I pass the label value to another file php file? How can I POST it?

The label and dropdown menu ;

<?php echo '<select name="type" id="category" onchange="changeOwner();">

        <option value="Staf DC">Staf DC</option>
        <option value="Admin">Admin</option>

    </select></th>'; 

echo "<td align='center'><label id='own'></label></td>";

javascript;

<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
function changeOwner()
{
var selname = $("#category option:selected").val();  
$.ajax({ url: "new_getdata.php",

    data: {"selname":selname},

    type: 'post',

    success: function(output) {
        $("#own").html(output);


    }

   });
}
window.onload =  changeOwner();
</script>   

new_getdata.php

if (isset($_POST['selname'])) { 
$selname = $_POST['selname'];
$query = "SELECT * FROM owner2 where type='$selname'";
$res = mysql_query($query);

   while ($rows = mysql_fetch_assoc($res)) {
   $name = $rows['owner'];


   echo $name;
}
}

The variable $name will replace the label value dynamically each time the dropdown menu changes. How do I send label value to another php file? Lets say I want to post it to register.php

Answer 1

$.ajax({ url: "new_getdata.php",
   data: {"selname":selname},
   type: 'post',
   success: function(output) {
       $("#own").html(output);

       $.ajax({
          data: {label: output},
          url: "regisster.php"
       });
   }
});