can i fetch and populate different fields using function(data) jquery? - php

Question

i want to populate two fields when some one fill up card no.
<input type="text" id="name" name="name" class="form-control" Value="<?php echo $grow['name']; ?>">
<input type="text" id="address" name="address" class="form-control" Value="<?php echo $grow['address']; ?>">
but this code populate field one by one. can any one suggest be better code for populating two fields from database. Thank you
jquery
<script type="text/javascript">
$(document).ready(function()
{
$("#krishi").keyup(function()
{
var k=$(this).val();
var q="name";
$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q='+q,
cache: false,
success: function(data)
{
if(data){
$("#name").val(data);
$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q=address',
cache: false,
success: function(data)
{
if(data){
$("#address").val(data);
}
}
});
}else
$("#name").val("");
$("#address").val("");
}
});
});
});
</script>
getresult.php
<?php
define('INCLUDE_CHECK',true);
include("mysql.php");
$k=$_POST['k'];
$q=$_POST['q'];
$sql=mysql_query("select * from inward where krishi='$k'");
$row=mysql_fetch_array($sql);
echo $row[$q];
?>

Answer 1

Try to extract both name and address from database and json them
$k=$_POST['k'];
$q=$_POST['q'];
$sql=mysql_query("select address,name from inward where krishi='$k'");
$row=mysql_fetch_array($sql);
$result = array(
'name'=>$row['name'],
'address'=>$row['address']);
echo json_encode($result);
After that parse them via jquery
$.ajax
({
type: "POST",
url: "getresult.php",
data: 'k='+k+'&q=address',
cache: false,
success: function(data)
{
if(data){
var parsedData = jQuery.parseJSON(data);
$("#name").val(parsedData.name);
$("#address").val(parsedData.address);
}
}
});

Answer 2

Javascript Code:
<script type="text/javascript">
$(document).ready(function()
{
$("#krishi").keyup(function()
{
var k = $(this).val();
var q = "name";
$.ajax({
type: 'POST',
url: "getresult.php",
data: 'k='+k,
cache: false,
success: function(data)
{
var jsonArr = $.parseJSON(data);
if(typeof response =='object')
{
$("#name").val(jsonArr.name);
$("#address").val(jsonArr.address);
}
else
{
$("#name").val("");
$("#address").val("");
}
}
});
});
});
</script>
PHP Code:
<?php
define('INCLUDE_CHECK',true);
include("mysql.php");
$k = $_POST['k'];
$sql = mysql_query("select * from inward where krishi='$k'");
$row = mysql_fetch_assoc($sql);
echo json_encode(array('name' => $row['name'], 'address' => $row['address']);
?>