Returning JSON array to ajax

Question

I'm getting stuck here.

I'm trying to trigger ajax when a div is selected and then get a JSON response. All I've been able to return is the error code, and I'm not sure why. Here's the simplified version (still getting same response as the actual). Maybe there's a better way? Thanks for your help.

jquery:

<html lang="en">
<meta charset="utf-8">
  <link rel="stylesheet" href="//code.jquery.com/ui/1.11.1/themes/smoothness/jquery-ui.css">
  <script src="//code.jquery.com/jquery-1.10.2.js"></script>
  <script src="//code.jquery.com/ui/1.11.1/jquery-ui.js"></script>
<style>
  #selectable .ui-selecting { background: #FECA40; }
  #selectable .ui-selected { background: #F39814; color: white; }
  #selectable { list-style-type: none; margin: 0; padding: 0; width: 60%; }
  #selectable li { margin: 3px; padding: 0.4em; font-size: 1.4em; height: 18px; }
</style>

<script>

  $(function(){


$("#selectable").selectable({
  selected : function(event, ui) {
    if($(ui.selected).hasClass('ui-widget-content')){
         var rptID = ($(ui.selected).val());
             $.ajax(
   {

      type: "POST",
      dataType: "json",
      url: "hello2.php",
      success: function( response ) 
      { 

        alert(1);
      },
      error: function( error )
      {
        console.log(error);


      }

   } );




    } 
    }
});

});
  </script>

  <body><ol id="selectable">
  <li class="ui-widget-content" value=1 >Item 1</li>
  <li class="ui-widget-content" value=2>Item 2</li>
  <li class="ui-widget-content" value=3>Item 3</li>
</ol></body>
</html>

php:

<?php
 $json = array('hello' => 'hello');
    header('Content-Type: application/json');
        echo json_encode($json); 
 ?>

Have you tried to remove the header line in the php file?

kiks73
– kiks73

2014-08-19 15:26:56 +00:00
Commented Aug 19, 2014 at 15:26 — kiks73
– kiks73, Commented Aug 19, 2014 at 15:26
alert(response): maybe?

AbraCadaver
– AbraCadaver

2014-08-19 15:27:11 +00:00
Commented Aug 19, 2014 at 15:27 — AbraCadaver
– AbraCadaver, Commented Aug 19, 2014 at 15:27
What error are you getting?

Barmar
– Barmar

2014-08-19 15:27:19 +00:00
Commented Aug 19, 2014 at 15:27 — Barmar
– Barmar, Commented Aug 19, 2014 at 15:27
Are there error in browser console? F12

SSA
– SSA

2014-08-19 15:27:26 +00:00
Commented Aug 19, 2014 at 15:27 — SSA
– SSA, Commented Aug 19, 2014 at 15:27
Doesn't make a difference.

yossarian_lives_next_door
– yossarian_lives_next_door

2014-08-19 15:27:53 +00:00
Commented Aug 19, 2014 at 15:27 — yossarian_lives_next_door
– yossarian_lives_next_door, Commented Aug 19, 2014 at 15:27

Marc B · Accepted Answer · 2014-08-19 15:36:50Z

3

php?> is going to cause a syntax error/warning at the PHP level, and that syntax error output will probably get tagged onto the end of your JSON output, causing a JSON parse error on the client.

PHP code blocks are simply <?php ... ?>.

e.g.

[marc ~]$ cat z.php
<?php
echo json_encode('foo');
php?>
[marc ~]$ php z.php
"foo"PHP Notice:  Use of undefined constant php - assumed 'php' in /home/marc/z.php on line 3

edit --- and now I see OP's edited away the php?> stuff...

answered Aug 19, 2014 at 15:36

Marc B

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1 Comment

Funk Forty Niner Over a year ago

I had made a comment to that effect (prior to your answer); OP responded afterwards saying it was a typo. I didn't even get a +1 for that and you got 2 lol - Now you need to edit your answer.

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