Question

I have this page that lists info from my mysql database. It requires an ID I $get from the url from the current page I’m on: localhost/index.php?page=id

I wanted to add a search function so that the content dynamicly changes, as you type.
Its functional but the ID that I require does not get passed when using the search function, only on the first include.

My code looks like this (I omitted some irrelevant code and html):

<?php
    $id = $_REQUEST['page']; 
?>

<script type="text/javascript">


$(document).ready(function() {


$("#faq_search_input").keyup(function()
{
var faq_search_input = $(this).val();
var dataString = 'keyword='+ faq_search_input;
if(faq_search_input.length>1)

{
$.ajax({
type: "GET",
url: "listing.php",
data: dataString,
beforeSend:  function() {


},
success: function(server_response)
{

$('#results').html(server_response).show();
$('span#faq_category_title').html(faq_search_input);

}
});
}return false;
});
});

</script>
...
...

    <form method="get" action="">
        <input  name="query" type="text" id="faq_search_input" />
    </form>
...
...
<td colspan="8">
        <ul id="sortable">
        <?php  require_once('listing.php');?>
        </ul>
</td>
...
...

Actually I think I’m loading the listing.php file twice now, as I see it. One time with the require_once and one time with my JavaScript. Can I improve this as well somehow?

So my question is: How can I get the ID to work with my javascript and could I restructer the code so I only load the listing.php once?

Please let me know if I am unclear, Im having a hard time explaining this.

Thanks a lot for any help

Answer 1

you should change dataString to

var dataString = 'keyword='+ faq_search_input +'&page=<?php echo $id; ?>';