Jquery ajax return null

Question

I am using the Jquery function,

function ajax_fun(id,val)
{
$.ajax({
    type: "GET",
    url: "data.php",
    dataType: "json",
    data: { pid: 1, cart_id: id, val: val },
    success: function (returndata){
       console.log(returndata);
    }  
});
}

Call

ajax_fun(1,100);

data.php

<?php
if(isset($_GET['pid']) && $_GET['pid']==1)
{

  $cart_id = $_GET['cart_id'];
  $val  = $_GET['val']+0.06;
  $arr = array('cart_id'=> $cart_id, 'total' => $val);
  json_encode($arr);
}
?>

Console returns 'null' . It didn't return cart_id and total.

Can anyone help?

.

You didn't echo the output..

Vishal
– Vishal

2012-08-09 06:49:29 +00:00
Commented Aug 9, 2012 at 6:49 — Vishal
– Vishal, Commented Aug 9, 2012 at 6:49

thecodeparadox · Accepted Answer · 2012-08-09 06:55:39Z

3

json_encode($arr);

shoule be

echo json_encode($arr);

You forget to echo the result.

Full code:

<?php
if(isset($_GET['pid']) && $_GET['pid']==1) {
  $cart_id = $_GET['cart_id'];
  $val  = $_GET['val']+0.06;
  $arr = array('cart_id'=> $cart_id, 'total' => $val);
  echo json_encode($arr); // write echo before json_encode($arr)
}
?>

edited Aug 9, 2012 at 6:55

answered Aug 9, 2012 at 6:48

thecodeparadox

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Comments

Rab Khan · Accepted Answer · 2012-08-09 06:49:59Z

1

RE-Write this line as...

echo json_encode($arr);

answered Aug 9, 2012 at 6:49

Rab Khan

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Comments

Muthu Kumaran · Accepted Answer · 2012-08-09 06:50:56Z

1

You need to add 'echo' before json_encode which return the data

<?php
if(isset($_GET['pid']) && $_GET['pid']==1)
{

  $cart_id = $_GET['cart_id'];
  $val  = $_GET['val']+0.06;
  $arr = array('cart_id'=> $cart_id, 'total' => $val);
  echo json_encode($arr);
}
?>

answered Aug 9, 2012 at 6:50

Muthu Kumaran

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