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

Carlos Lantigua
Carlos Lantigua
5,938 Points
Posted by Carlos Lantigua
Carlos Lantigua
5,938 Points

AJAX basics Stage 4 challenge Code not working :,,, (

When I try to run this I get the following error message in the console Failed to load resource: the server responded with a status of 404 ()

when I click on the link it says that it's a bad link. But I don't see any issues with my API link. 

$(document).ready(function() {

 $('form').submit(function (evt) {
   evt.preventDefault();
    var $searchField = $('#search');

   //the AJAX part
   var flickerAPI = "http://api.flickr.com/services/feeds/photo_public.gne?jsoncallback=?";
   var animal = $searchField.val();
    var flickrOptions = {
      tags: animal,
      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, flickrOptions, displayPhotos);

  }); // end click

}); // end ready

1 Answer

Steven Parker
Steven Parker
229,644 Points
Steven Parker
Steven Parker
229,644 Points

It looks like you have a typo in your URL. You have "photo_public.gne" instead of "photos_public.gne" (with an "s").

Carlos Lantigua
Carlos Lantigua
5,938 Points
Carlos Lantigua
Carlos Lantigua
5,938 Points

Thanks, I’ve come to realize one of the hardest parts of coding is type-o haha