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

Tushar Singh
Tushar Singh
Courses Plus Student 8,692 Points

ajax part 4 challenge

I am not sure why my code is not working!!

html file

<!DOCTYPE html>
  <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>
  <div class="grid-container centered">
    <div class="grid-100">
      <div class="contained">
        <div class="grid-100">
          <div class="heading">
            <h1>Flickr Photo Search</h1>
              <label for="search">Type a search term</label>
              <input type="search" name="search" id="search">
              <input type="submit" value="Search" id="submit">


        <ul id="photos">


And javascript file

$(document).ready(function() {

 $('form').submit(function (evt) {
    var $searchTerm = $("#search"); 

    // the AJAX part
    var flickerAPI = "http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?";
    var animal = $searchTerm.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>';
    $.getJSON(flickerAPI, flickrOptions, displayPhotos);

  }); // end click

}); // end ready

1 Answer

Toni Ojala
Toni Ojala
17,570 Points

I didn't see anything wrong with your code and literally copied and pasted it to a workspace of mine and it worked fine minus the css I'm lacking. You might want to check whether your javascript file is actually named app.js and is in the js folder.