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Start your free trialDongyun Cho
2,256 Pointsif(photo == null) ?
'''javascript $(document).ready(function() {
$('form').submit(function (evt) { evt.preventDefault(); var $searchField = $('#search'); var $submitButton = $('#submit'); $searchField.prop("disabled", true); $submitButton.prop("disabled", true).val("searching...");
var flickerAPI = "http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?";
var term = $searchField.val();
var flickrOptions = {
tags: term,
format: "json"
};
function displayPhotos(data) {
var photoHTML = '<ul>';
$.each(data.items,function(i,photo) {
if(photo == null){
photoHTML += "<p>NO pics</p>");
} else {
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);
$searchField.prop("disabled", false);
$submitButton.attr("disabled", false).val("Search");
}
$.getJSON(flickerAPI, flickrOptions, displayPhotos); }); // end click
}); // end ready '''
I added some other code as the teacher said at the end of this video. When some picture I wrote in the form is not exist in Flickr, the browser should show us a text saying, there's not a picture that I want. The text "No pics", however, does not turn out. what should I do to fix it?
3 Answers
Tyler Corum
3,514 PointsHere's what I want you to do to gain a little insight into your program:
Right after this line
function displayPhotos(data) {
put
console.log(data);
Then do a search for something weird like alkadfgh
and see where that insight gets you!
Best of luck!
barnetobeka
Courses Plus Student 10,895 Points$(document).ready(function() {
$('form').submit(function (event) { event.preventDefault(); var flickerAPI = "http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?"; var animal = $("#search").val(); var flickrOptions = { tags: animal, format: "json" }; function displayPhotos(data) { var photoHTML=""; if($.isEmptyObject(data.items) === true){ photoHTML += '<li class="newList">Search not Found</li>'; }else{ $.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 }; $("#photos").html(photoHTML); }; // diasplay photos $.getJSON(flickerAPI, flickrOptions, displayPhotos); }); // end submit
}); // end ready
barnetobeka
Courses Plus Student 10,895 PointsUse the Jquery selector of $.isEmptyObject();
Tyler Corum
3,514 PointsTyler Corum
3,514 PointsIf you still get stuck (hint: you'll discover by clicking the triangle, data.items... length: 0).
I include 2 different working solutions of the displayPhotos() function in question. I looked in two places for reference (bookmark these two repositories, developer.mozilla.org and jQuery.com!) http://api.jquery.com/jquery.each/ https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/prototype
Both solutions are testing working. I just commented out one of them below.