Saturday, 17 August 2013

Load Ajax response div into three different div

Load Ajax response div into three different div

I am working on ajax, the ajax function is working fine but the problem is
to break ajax response between three div. Let me explain in detail.
1. I have a page where i have one select box and three input , and by the
onchange event i want the ajax value in input boxes.
for further explaination here is my code(Ajax Response):
<?php
include("include/config.php");
$enrol=$_GET['q'];
$query=mysqli_query($con,"select * from dbo_tbmembermaster where
vc_enrol_no='$enrol'");
$result=mysqli_fetch_assoc($query);
$mem_id=$result['nu_member_id'];
$result['nu_member_id'];
$query2=mysqli_query($con,"select SUM(nu_amount) as amt_sub from
dbo_tbmemberdues where nu_member_id='$mem_id' and
vc_due_for='Subscription'");
$result2=mysqli_fetch_assoc($query2);
echo '<div id="d1">';
echo $amt_sub=$result2['amt_sub'];
echo '</div>';
$query3=mysqli_query($con,"select SUM(nu_amount) as amt_fund from
dbo_tbmemberdues where nu_member_id='$mem_id' and vc_due_for='Employees
Welfare Fund'");
$result3=mysqli_fetch_assoc($query3);
echo '<div id="d2">';
echo $amt_fund=$result3['amt_fund'];
echo '</div>';
$query4=mysqli_query($con,"select SUM(nu_amount) as amt_fund from
dbo_tbmemberdues where nu_member_id='$mem_id' and vc_due_for='Cause
List'");
$result4=mysqli_fetch_assoc($query4);
echo '<div id="d3">';
echo $amount=$result4['amt_fund'];
echo '</div>';
?>
Now this is my response and please note here i have some values in three
different div <div id="d1">,<div id="d2">,<div id="d3">
2. Now Iam getting this response on my page by the Id <div
id="result"></div>. So After Ajax call I have d1,d2,d3 on my page and the
value of these div I am trying to get into three input box. I am doing
this by javascript as:
function showdues()
{
var aa=document.getElementById("d1").innerHTML;
//alert("vhsgfhsg");
document.getElementById("sub_due").value=aa;
}
and I am calling ajax function and this function by onchange like this
<select name=dues onchange="Javascript:
showUser(dues.value);showdues();">, but ajax function is working but this
not working and I am not getting anything into input box.
It may be a weird idea to do this but I uses my best and your any type of
idea , what I am missing and what should I have to do.. If there is any
better idea Plz Let me know....
thanks in advance.

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