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Home/ Questions/Q 8346011
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T06:52:14+00:00 2026-06-09T06:52:14+00:00

I am using the Jquery function, function ajax_fun(id,val) { $.ajax({ type: GET, url: data.php,

  • 0

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?

.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T06:52:15+00:00Added an answer on June 9, 2026 at 6:52 am
    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)
    }
    ?>
    
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