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JavaScript AJAX Basics (retiring) AJAX and APIs Stage 4 Challenge Answer

Sharina V. Jones
Sharina V. Jones
11,459 Points
Posted by Sharina V. Jones
Sharina V. Jones
11,459 Points

AJAX stage 4 challenge

I'm trying to complete the Stage 4 challenge but my code doesn't work. I'm not getting any error messages, but I'm not getting any pictures either.

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>AJAX Flickr Photo Search</title>
  <link href='http://fonts.googleapis.com/css?family=Varela+Round' rel='stylesheet' type='text/css'>
  <link rel="stylesheet" href="css/main.css">
  <script src="http://code.jquery.com/jquery-1.11.0.min.js"></script>
  <script src="js/app.js"></script>
</head>
<body>
  <div class="grid-container centered">
    <div class="grid-100">
      <div class="contained">
        <div class="grid-100">
          <div class="heading">
            <h1>Flickr Photo Search</h1>
            <form>
              <label for="search">Type a search term</label>
              <input type="search" name="search" id="search">
              <input type="submit" value="Search" id="submit">
            </form>

          </div>
        </div>

        <ul id="photos">

        </ul>
      </div>
    </div>
  </div>
</body>
</html>

$(document).ready(function() {
  $('form').submit(function(evt) {
    evt.preventDefault();    
    var flickerAPI = "http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?";
    var input = ('#search').val();
    var flickrOptions = {
      tags: input,
      format: "json"
    };

    function displayPhotos(data) {
      var photoHTML = '<ul>';
      $.each(data.items,function(i,photo) {
        photoHTML += '<li class="grid-25 tablet-grid-50">';
        photoHTML += '<a href="' + photo.link + '" class="image">';
        photoHTML += '<img src="' + photo.media.m + '"></a></li>';
      }); // end each
      photoHTML += '</ul>';
      $('#photos').html(photoHTML);
    }
    $.getJSON(flickerAPI, input, displayPhotos);
  });//end submit

});

3 Answers

Brandon Dyal
Brandon Dyal
16,544 Points
Brandon Dyal
Brandon Dyal
16,544 Points

It looks like you're missing the jquery selector $ in front of ('#search').val(); . Try var input = $("#search").val();

Marcus Parsons
Marcus Parsons
15,719 Points

Good catch. Also, in the getJSON method, "input" should be "flickrOptions".

Marked as best answer

Ben Ghabili
Ben Ghabili
7,840 Points
Ben Ghabili
Ben Ghabili
7,840 Points

I had a similar problem getting result when using Launchpad. Same code perfectly works when running in local machine such as your own laptop or desktop.

Marcus Parsons
Marcus Parsons
15,719 Points

This code would not work perfectly as it is as there are a couple errors present.

Max Weston
Max Weston
11,432 Points
Max Weston
Max Weston
11,432 Points

I think flickr have changed their feeds url to this as mine didn't work before I changed it to this:

var flickerAPI = "https://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?";

hope this works.

Marcus Parsons
Marcus Parsons
15,719 Points

That is the same URL as used above and the one used above is still the valid public feed URL.

Marcus Parsons
Marcus Parsons
15,719 Points

That is the same URL as used above and the one used above is still the valid public feed URL.