Jquery ajax return nothing from requested php page

Question

I'm sending request by $.ajax method from test.php to ajax.php page. The console shows "200 ok" that means the request is ok. But return nothing, although the console doesn't show any error. My used pages are as following:

test.php

<div id="return-data">
    <ul class="return-lists">
    </ul>
</div>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
$('.submit-form').click(function() {
    var name = "tanvir";
    var address = "Dhaka";

    var cData = "name=" + encodeURIComponent(name) + "&address=" + encodeURIComponent(address);

    $.ajax({
        url: "ajax.php",
        type: "POST",
        data: cData,
        success: function(data)
        {
            alert(data);
            $('#return-data .return-lists').append(data);
        }
    });
});
</script>

And ajax.php

<?php
    if( isset($_POST['name']) ) {
    $name = $_POST['name'];
    $address = $_POST['address'];

    $lists = '';
    $lists .= '<li>' . $name . '</li>';
    $lists .= '<li>' . $address . '</li>';


    return $lists;
    exit;
}

I also have tried by removing if( isset($_POST['name']) ) {} and by using echo instead of return in ajax.php

Yeah, it is a button. And on click it, the request is submitted. — Sobuj alice
– Sobuj alice, Commented Aug 21, 2015 at 18:18
If you check the network tab you will see two requests. You have to stop the button from performing its default behaviour - namely submitting the form. api.jquery.com/event.preventdefault — Andreas
– Andreas, Commented Aug 21, 2015 at 18:21

HenryDev · Accepted Answer · 2015-08-21 18:28:42Z

1

Try this one. Change return $list for echo $list and no need to use exit at the end of ajax.php

 <?php
 if( isset($_POST['name']) ) {
  $name = $_POST['name'];
  $address = $_POST['address'];

  $lists = '';
  $lists .= '<li>' . $name . '</li>';
  $lists .= '<li>' . $address . '</li>';


 echo $lists;

 }

answered Aug 21, 2015 at 18:28

HenryDev

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1 Comment

Sobuj alice Over a year ago

Thanks henry, I tried to accept your answer immediately. But on my account your answer was disappeared after displaying for few minutes. Thanks again.

Sobuj alice · Accepted Answer · 2015-08-21 18:25:39Z

0

Yeah, I have solved the problem by normally putting "echo" instead of "return". Because the "return" return an string and echo print an string. Thanks to all.

answered Aug 21, 2015 at 18:25

Sobuj alice

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