Jquery :: Ajax Validation - How To Blank Out Field In Form

Feb 28, 2011

I have a scenario where I have the following jquery
Code:
$("#frompdc<?php echo $JavaCnt;?>").change(function(){
var id = $('#id<?php echo $JavaCnt;?>').attr('value');
var frompdc = $('#frompdc<?php echo $JavaCnt;?>').attr('value');
var topdc = $('#topdc<?php echo $JavaCnt;?>').attr('value');
$.ajax({
type: "POST",
url: "AJ_Update.php",
data: "firm=y&id="+ id + "&frompdc=" + frompdc&topdc=" + topdc
});

And some php to insert this data
Code:
mysql_query("UPDATE TBLTRANSFERS SET FROMPDC = ".$_POST['frompdc'].", MOD_TS = '". $timenow ."' WHERE ID = ".$_POST['id']);

I'm trying to throw some validation in here which I can do in php (a language that I'm much more proficient in) but can't figure out how to get the two together. I want to be able to do a select against another table I have for both the frompdc and the topdc.
Code:
$tofpdcresult = mysql_query("SELECT * FROM tbltransferspdcexclusions where FROMPDC = '".$frompdc."' AND TOPDC = '".$topdc."'");
$tofpdcnum_rows = mysql_num_rows($tofpdcresult);
And if it exists I want to give a popup message to the user as well as blank out the field.

I'm struggling with two things:
How do I get ajax to run two queries and return the results to a popup window?
How do I blank out the field in the form?

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