I want to fetch value through jquery Ajax in PHP. but when I run below this code value is fetch properly but all value are joined together when fetching and together value print all field, but I want to print the separate value in the separate field. my code is below, please solve this problem
index.php
<!DOCTYPE html>
<html>
<head>
<title>Untitled Document</title>
</head>
<body>
<select name="Pro" onChange="my_validate_func()" id="pro_serv">
<option value=""></option>
<option value="" ><font size="-1">Add New Product</font></option>
<option value="Development"><font size="-1">Development</font></option>
<option value="Maintance"><font size="-1">Maintance</font></option>
</select>
<br />
<input type="text" name="desc" id="desc" value="" />
<br />
<input type="text" name="unitprice" id="unitprice" value="" />
<br /><input type="text" name="discount" />
<script type="text/javascript" src="bootstrap/jquery1.8.3jquery.min.js">
</script>
<script>
$( document ).ready(function() {});
function my_validate_func() {
var pro_serv = $('#pro_serv').val();
if ($('#pro_serv').val() != "" ) {
$.ajax({
type: "POST",
url: 'check.php',
data: { pro_serv: pro_serv},
success: function(response) {
$('#desc').val(response);
$('#unitprice').val(response);
$('#tax').val(response);
}
});
}
}
</script>
</body>
</html>
check.php
<?php
include('database/db.php');
$type=$_POST['pro_serv'];
$sql="Select * from product_service_details where type='$type'";
$result=mysql_query($sql);
if($dtset=mysql_fetch_array($result))
{
$desc=$dtset['desc'];
$unitprice=$dtset['unitprice'];
$tax=$dtset['tax'];
echo $desc;
echo $unitprice;
echo $tax;
}
?>
