jquery ajax request returns undefined

Question

I am trying to have a link perform a really simple, request, but I can't seem to figure out why what I returned is "undefined".

Currently the link does a manual page request, but it's such a simple subscription thing, I want it to not refresh the page (hence the preventDefault). But if the JS doesn't work or it's blocked, the link should do a normal page request (best of both worlds?).

Here is the JS that captures the click on a link

  $('#subscribe-link').click(function(e)
  {
    e.preventDefault();
    var type = $(this).data('sub');
    $.post('/includes/ajax/subscribe-article.php', { 'type':type },
    function(data)
    {
      alert(data.result);
      if (data.result == 'subscribed')
      {
        alert('subscribe');
      }
      else if (data.result == 'unsubscribed')
      {
        alert('unsubscribe');
      }
    });
  });

And here is the PHP that feeds it:

if($_POST && isset($_SESSION['user_id']) && $_SESSION['user_id'] != 0)
{
  if (isset($_POST['type']))
  {
    if ($_POST['type'] == 'subscribe')
    {
      echo json_encode(array("result" => "subscribed"));
      return;
    }

    if ($_POST['type'] == 'unsubscribe')
    {
      echo json_encode(array("result" => "unsubscribed"));
      return;
    }
  }
}

Now, I've checked what "data" returns by itself which is this:

{"result":"unsubscribed"}

Which is correct, I'm not sure what I'm missing this time.

Florian Heer · Accepted Answer · 2016-11-01 21:35:50Z

2

As the variable data contains the JSON representation of your expected result, it is plainly a String. Yet you try and access information contained in that string as an object. For this to work, you first need to create an object from the string by decoding the returned JSON:

var myData = JSON.parse(data);
alert(myData.result);
...

edited Nov 1, 2016 at 21:35

answered Nov 1, 2016 at 21:29

Florian Heer

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2 Comments

Jakub Judas Over a year ago

dataType: "json" might also help

NaughtySquid Over a year ago

Ah interesting thanks, I was previously using this exact same method, but returning numbers instead of text, so I guess it's picked up differently.

mlewis54 · Accepted Answer · 2016-11-01 21:28:41Z

0

You need to parse the result as JSON. IE, data_array=JSON.parse(data);

answered Nov 1, 2016 at 21:28

mlewis54

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