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Martin Coutts
18,154 PointsPhotos not displaying, just blank links
My photos won't display, I have a lot of empty links where the pictures should be. Here is a copy of my js code:
$(document).ready(function(){
$('button').click(function() {
$("button").removeClass("selected");
$(this).addClass("selected");
var flickerAPI = "http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?";
var animal = $(this).text();
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.n + '"></a></li>';
});
photoHTML += '</ul>';
$('#photos').html(photoHTML);
}
$.getJSON(flickerAPI, flickrOptions, displayPhotos);
});
}); //end ready
2 Answers
Sean Urgel
Courses Plus Student 6,650 PointsThe $.getJSON(1, 2, displayPhotos) function is inside function displayPhotos(). You should call it outside the function.
Rhys Kearns
4,976 Points$(document).ready(function(){
$('button').click(function() {
$("button").removeClass("selected");
$(this).addClass("selected");
var flickerAPI = "http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?";
var animal = $(this).text();
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.n + '"></a></li>'; });
photoHTML += '</ul>'; $('#photos').html(photoHTML); }
$.getJSON(flickerAPI, flickrOptions, displayPhotos);
});
}); //end ready
I made your code a little more readable in the forums incase anyone wants to help