Slider - How would you make a simple slider? - javascript

EDIT: I WANT MY SLIDER TO BE EASY TO YOU AND ONCE A LINK IS CLICKED THE IMAGE THAT CORRELATES WITH LINK SHOWS.
I'm looking to make a really simple "slider" that if you click a link, the img shows that correlates with it. I've been trying to find something for a bit now and things are either too flash or don't suit my needs. This came close: http://jsfiddle.net/bretmorris/ULNa2/7/
I would something a little simpler that can be applied easily to multiple images for different divs.
This is what my code looks like with just a plain img tag to it:
<div id="adobe_content" class="hdiv"><button class="x">X</button>
<img src="images/adobefront.png"><br>
<img src="images/adobeinside.png"><br>
<img src="images/adobeback.png"><br>
<h5>Adobe Brochure</h5>
<p>
I wanted to make something functional and out the box that Adobe might consider giving out. It's clean like their work and sparks interest in opening the brochure with the cut out in the center. The front flap is able to slide into a slot on the right side for a neat logo. They're now more interested in their cloud, but the information is inside is still relevant!
</p>
<b>Programs used: Adobe Illustrator, InDesign and Photoshop.</b>
</div>
The code doesn't work for me because, well I partially don't understand it, and I'm not sure how to make it suit my needs (especially if I got up to multiple images) like correlating with an image.

Perhaps understanding what is going on would maybe get you on the right track, so here is an explanation:
$('a.prev').click(function() {
prevSlide($(this).parents('.slideshow').find('.slides'));
});
//clicking image goes to next slide
$('a.next, .slideshow img').click(function() {
nextSlide($(this).parents('.slideshow').find('.slides'));
});
This part is relatively straightforward, when you click on the previous or next links, call the prevSlide or nextSlide function, passing the collection of slides as an argument.
//initialize show
iniShow();
function iniShow() {
//show first image
$('.slideshow').each(function() {
$(this).find('img:first').fadeIn(500);
})
}
Initialize the slideshow by finding each slideshow on the page and fading in the first image. $(this) refers to the <div class="slideshow"> parent, find all child image tags and take the first, fade that element in (and do it in 500 milliseconds).
function prevSlide($slides) {
$slides.find('img:last').prependTo($slides);
showSlide($slides);
}
function nextSlide($slides) {
$slides.find('img:first').appendTo($slides);
showSlide($slides);
}
The prevSlide and nextSlide functions both rearrange the order of images, this line in particular:
$slides.find('img:first').appendTo($slides);
Is moving the first image to the end of the images, so:
<img src="http://placekitten.com/300/500" width="300" height="500" />
<img src="http://placekitten.com/200/400" width="200" height="400" />
<img src="http://placekitten.com/400/400" width="500" height="400" />
becomes:
<img src="http://placekitten.com/200/400" width="200" height="400" />
<img src="http://placekitten.com/400/400" width="500" height="400" />
<img src="http://placekitten.com/300/500" width="300" height="500" />
$slides.find('img:last').prependTo($slides); does the inverse and moves the last image to the beginning.
function showSlide($slides) {
//hide (reset) all slides
$slides.find('img').hide();
//fade in next slide
$slides.find('img:first').fadeIn(500);
}
Finally, showSlide accepts the collection of images, hides all of them and then fades in the first image (since the collection is reordered each time, the first is a different image).
Now, if you want a link for each image that will display a corresponding image, you could do something as simple as:
<a class="image" data-src="http://placekitten.com/300/500">Kitten 1</a>
<a class="image" data-src="http://placekitten.com/200/400">Kitten 2</a>
<a class="image" data-src="http://placekitten.com/400/500">Kitten 3</a>
<div id="image-container">
<img src="http://placekitten.com/300/500" />
</div>
and something like the following:
$('.image').on('click', function() {
var imageSrc = $(this).data('src');
$('#image-container img').prop('src', imageSrc);
});
Which will update the child image tag of <div id="image-container"> with the data-src attribute value in the clicked link.
http://jsfiddle.net/9sxt6n0t/
Hope this helps.

just a quick function to slide
function slideIt(images , prev , next){
$('.slideshow img:nth-child(1)').show();
var imagesLength = $(images).length;
var i = 1;
$(prev).click(function() {
$(images).hide();
if(i !== 1){
$(images + ':nth-child('+ (i - 1) +')').show();
i--;
}else{
$(images +':nth-child('+imagesLength +')').show();
i = imagesLength;
}
});
//clicking image goes to next slide
$(''+next+','+images+'').on('click',function() {
$(images).hide();
if(i !== imagesLength){
$(images + ':nth-child('+ (i + 1) +')').show();
i++;
}else{
$(images + ':nth-child(1)').show();
i = 1;
}
});
}
and use like that slideIt(Images , prevArrow , nextArrow)
slideIt('.slideshow img','a.prev' , 'a.next');
DEMO HERE

Related

Reload a DIV on click of another DIV

I have a div called masterdiv, inside this div there are 3 other div div1, div2, and div3,
This is the html for these html:
<div id="masterdiv" class="masterdivclass">
<div id="div1"><img class="div1class" src="image1.jpg" id="div1id" /></div>
<div id="div2"><img class="div2class" src="image2.jpg" id="div2id" /></div>
<div id="div3"><img class="div3class" src="image3.jpg" id="div3id" /></div>
</div>
I also have another div:
<div id=”reload”><img src="reload.png" width="200" height="70" onclick=loadDIV();></div>
What I’m trying to do is to reload the masterdiv div whenever the reload div is clicked on. Hiding and then showing the div isn’t enough as I need the content to be reloaded when the refresh div is clicked on. I don’t want to reload the entire page, just the masterdiv which contains the 3 other div. But I’m not certain this is possible.
I’m trying to do it with this Javascript function:
<script type="text/javascript">
function loadDiv(){
$("<div id="masterdiv" class="masterdivclass">
<div id="div1"><img class="div1class" src="image1.jpg" id="div1id" /></div>
<div id="div2"><img class="div2class" src="image2.jpg" id="div2id" /></div>
<div id="div3"><img class="div3class" src="image3.jpg" id="div3id" /></div>
</div>").appendTo("body");
}
</script>
This isn’t working, I think maybe I'm going about this in the wrong way? Maybe I’m missing something very simple here? I’d really appreciate any help with this, thank you in advance!
UPDATE
After reconsidering my project's requirements, I need to change part of my question, I now need to randomise the images displayed in the divs, and have a new random image load every time the reload div is clicked on. I also need to remove each class that’s currently in each of the three divs and then reattach the same classes to the divs (if I don’t remove and reattach the classes then the divs just display the plain images without any class/effect applied to them, it seems like I need to reload the class every time I load an image into a div in order for the class/effect to be applied successfully).
I have 5 images, and I’m using each div’s id tag to attach a random image to each div.
First I’m assigning the 5 different images to 5 different ids:
<script>
document.getElementById('sample1').src="images/00001.jpg";
document.getElementById('sample2').src="images/00002.jpg";
document.getElementById('sample3').src="images/00003.jpg";
document.getElementById('sample4').src="images/00004.jpg";
document.getElementById('sample5').src="images/00005.jpg";
</script>
And then I’m trying to use the following Javascript to load a randomised id (and its assigned image) to each of the 3 divs when the reload div is clicked:
<script>
$(function() {
$('#reload').on('click',function(){
$("#masterdiv").find("div[id^='div']").each(function(index){
//First, remove and reattach classes “div1class”, “div2class” and “div3class”
//from “easyDIV”, “mediumDIV” and “hardDIV” respectively:
$(“#easyDIV”).removeClass('div1class');
$(“#easyDIV”).addClass('div1class');
$(“#mediumDIV”).removeClass('div2class');
$(“#mediumDIV”).addClass('div2class');
$(“#hardDIV”).removeClass('div3class');
$(“#hardDIV”).addClass('div3class');
//Get a random number between 1 and 5, then attach it to “sample”,
//so that the result will be either “sample1”, “sample2”, “sample3”, “sample4” or “sample5”,
//call this variable “variablesample”:
var num = Math.floor(Math.random() * 5 + 1);
variablesample = "sample" +num;
//Attach this randomised id to all three divs using “variablesample”:
jQuery(this).prev("easyDIV").attr("id",variablesample);
jQuery(this).prev("mediumDIV").attr("id",variablesample);
jQuery(this).prev("hardDIV").attr("id",variablesample);
});
var p = $("#masterdiv").parent();
var el = $("#masterdiv").detach();
p.append(el);
});
});
</script>
I’m trying to make it so that all 3 divs will show the same randomised picture (that’s why they’re all sharing the variable “variablesample”), and each div will reload its own class/effect (div1class, div2class and div3class) but it’s not working. I’m not sure if it’s correct to use jQuery inside a Javascript function, or if my syntax for updating the ids of the divs is incorrect.
Perhaps my logic to solving this problem is all wrong? I’d really appreciate any more help with this problem. Thanks again in advance!
Original question was edited many times, so here is the correct answer for the latest edit. Answer to the question; "How to use random image, but same image on all 3, and 3 class on/off switching":
$(function() {
var imageArray = [
'https://via.placeholder.com/40x40',
'https://via.placeholder.com/80x40',
'https://via.placeholder.com/120x40',
'https://via.placeholder.com/160x40',
'https://via.placeholder.com/200x40'];
reloadImages(imageArray);
$('#reload').on('click',function(){
$( "#masterdiv img[id^='div']" ).each(function(index){
$(this).removeClass("div"+(index+1)+"class");
$(this).fadeOut( "slow", function() {
if(index==0) {
reloadImages(imageArray);
}
$(this).addClass("div"+(index+1)+"class");
$(this).fadeIn();
});
});
});
});
function shuffleArray(array) {
for (var i = array.length - 1; i > 0; i--) {
var j = Math.floor(Math.random() * (i + 1));
var temp = array[i];
array[i] = array[j];
array[j] = temp;
}
return array;
}
function reloadImages(array){
shuffleArray(array);
for(var i=0;i<3;i++){
// places the first image into all divs, change 0 to i if you want different images in each div
document.getElementById('div'+(i+1)+'id').src=array[0];
}
}
.div1class {
border:2px dashed #0F0;
}
.div2class {
border:2px dashed yellow;
}
.div3class {
border:2px dashed red;
}
#reload {
background-color:blue;
color:white;
width:100px;
height:30px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id='reload'>Click here</div>
<div id="masterdiv" class="masterdivclass">
<div id="div1">
<img src="https://via.placeholder.com/20x40" class="div1class" id="div1id" />
</div>
<div id="div2">
<img src="https://via.placeholder.com/20x40" class="div2class" id="div2id" />
</div>
<div id="div3">
<img src="https://via.placeholder.com/20x40" class="div3class" id="div3id" />
</div>
</div>
Line breaks and un-escaped quotes are why the functions is not working.
function loadDiv(){
$('#masterdiv').remove();
$("<div id='masterdiv' class='masterdivclass'><div id='div1'><img class='div1class' src='image1.jpg' id='div1id' /></div><div id='div2'><img class='div2class' src='image2.jpg' id='div2id' /></div><div id='div3'><img class='div3class' src='image3.jpg' id='div3id' /></div></div>").appendTo("body");
}
try:
function loadDiv(){
$("#masterdiv").load(location.href + " #masterdiv");
}
Here's the code pen demo:
http://codepen.io/anon/pen/xwgRWm
If content of container is not being changed dynamically then there is no point reloading it. appendTo wiil append DOM in existing DOM structure, you will need html() here which will replace the content inside container. Also note you had typo here onclick=loadDiv();
HTML:
<div id="masterdiv" class="masterdivclass">
<div id="div1"><img class="div1class" src="image1.jpg" id="div1id"/></div>
<div id="div2"><img class="div2class" src="image2.jpg" id="div2id"/></div>
<div id="div3"><img class="div3class" src="image3.jpg" id="div3id"/></div>
</div>
<div id="reload"><img src="reload.png" width="200" height="70" onclick=loadDiv();></div>
JS:
function loadDiv() {
$("#masterdiv").html('<div id="div1"><img class="div1class" src="image1.jpg" id="div1id" /></div>\
<div id="div2"><img class="div2class" src="image2.jpg" id="div2id" /></div>\
<div id="div3"><img class="div3class" src="image3.jpg" id="div3id" /></div>');
}

How to use jquery to swap the position of multiple images on click

I'm a novice at jquery/javascript, so apologies in advance if this is a dumb question.
I have 3 images and I want to be able to change their position when a user clicks either the second or the third one. For example, if the user clicks image #2, I want image #2 to become image #1 (left-most), then image #3 becomes image #2 (middle) and finally image #1 becomes image #3 (right-most). If the user clicks image #3, I want image #3 to become image #1 (left-most), then image #1 becomes image #2 (middle) and finally image #2 becomes image #3 (right-most).
If the user clicks the first image, I don't want anything to happen.
Here is my code:
<!DOCTYPE html>
<html>
<head>
<title>ImageSwap</title>
<script type="text/javascript" src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">
$(document).ready(function () {
var img = new Array("http://i.imgur.com/QWso5yB.png", "http://i.imgur.com/O4X3egC.png", "http://i.imgur.com/w3p2qLp.png");
$('#first').bind('click', firstChannel);
$('#second').bind('click', secondChannel);
});
function firstChannel() {
var tempImg = img[0];
img[1] = img[2];
img[2] = tempImg;
$('#home').attr('src',img[0]);
$('#first').attr('src',img[1]);
$('#second').attr('src',img[2]);
};
function secondChannel() {
var tempImg = img[0];
img[0] = img[2];
img[2] = img[1];
img[1] = tempImg;
$('#home').attr('src',img[0]);
$('#first').attr('src',img[1]);
$('#second').attr('src',img[2]);
};
</script>
</head>
<body>
<img id="home" src="http://i.imgur.com/QWso5yB.png">
<img id="first" src="http://i.imgur.com/O4X3egC.png">
<img id="second" src="http://i.imgur.com/w3p2qLp.png">
</body>
</html>
When clicking the 2nd and 3rd images, nothing happens. What the heck am I doing wrong here? Is there an easier way to do this? I've been pulling my hair out searching everywhere but can't seem to find an answer. Many thanks in advance for any feedback you can give me...
Why not make it simple? jsBin demo
A parent element:
<div id="channels">
<img src="http://i.imgur.com/QWso5yB.png">
<img src="http://i.imgur.com/O4X3egC.png">
<img src="http://i.imgur.com/w3p2qLp.png">
</div>
and some prepend():
$(function () {
var $chn = $("#channels");
$chn.on("click","img", function(){
$chn.prepend( this );
});
});
I mean, if the first image represents the current channel, than all you need to do (as above) is to prepend the clicked element. To style the first element simply use CSS img:first-child
EDIT: KEEP ORDER
jsBin demo with Order
If appending creates at some point a mess of channels order,
I'd suggest you to:
Create a MAIN or currently watching big image and place all channels inside a parent:
<img id="current">
<div id="channels">
<img src="//placehold.it/90x40/a7b&text=1">
<img src="//placehold.it/90x40/ba7&text=2">
<img src="//placehold.it/90x40/7ba&text=3">
<img src="//placehold.it/90x40/bb7&text=4">
<img src="//placehold.it/90x40/77a&text=5">
<img src="//placehold.it/90x40/ab7&text=6">
</div>
Than on a channel-click, set the clicked image src to the BIG image, and hide the clicked one:
$(function () {
var $img = $("#channels").find("img");
var $current = $("#current"); // The big image
$img.on("click", function(){
$current[0].src = this.src;
$img.show();
$(this).hide();
}).eq(0).click();
});
Example with a better UI

Html in-page hidden text that isn't shown until link is clicked

sort of new to html. I'm looking to create animation that when am image is clicked, it plays an animation that splits open a half page of text and in stuff. Something like this: http://sketchtoy.com/62368639
If you want to do things like that, you should really have a look at a javascript framework like jquery (www.jquery.com)
I find this one particularly easy to learn.
For what you want to do:
<div style="display: none;" id="mytext">Your text</div>
<a onclick="$('#mytext').show()"></a>
The basic process is
Add the click handler for the images
Find the last image in the row that the clicked image is in
set the contents of the expanding element
insert the expanding element after the image from step 2
show the expanding element (in this case a slideDown animation)
This is using jQuery library, which you did not tag, but it makes doing this a lot easier. If you need a vanilla javascript approach one can be added.
HTML
<div id="container">
<img src="http://placehold.it/128x128" />
<img src="http://placehold.it/128x128" />
<img src="http://placehold.it/128x128" />
<img src="http://placehold.it/128x64" />
<img src="http://placehold.it/128x64" />
<img src="http://placehold.it/128x64" />
<div id="expander">
<img src="" />
<div id="info">
some info
</div>
</div>
</div>
JS
//Just making the expander half the height of the viewport
var winheight = $(window).height();
$("#expander").css("height",winheight/2);
$("img").click(function(){
var img = $(this);
var src = img.attr("src");
var afterImg = findLastImgInRow(img);
var expander = $("#expander");
//Hide the expander if previously open
expander.hide(0);
//just setting the insides
expander.find("img").attr("src",src);
expander.find("#info").html("This is a test");
//Put the expander after the last image in the row
//so it will appear between its row and the next
expander.insertAfter(afterImg);
expander.slideDown(600);
//This scrolls the page so that it will make the
//expander appear in the middle of the page
$('html, body').animate({
scrollTop: expander.offset().top-(winheight/4)
}, 600);
});
//Function to find the last image
//in a row by comparing their offset top values
function findLastImgInRow(img){
var imgTop = img.offset().top;
var img2 = img;
do{
if( img2.offset().top != imgTop ){
img2 = img2.prev();
break;
}
}while(img2=img2.next());
return img2;
}
JSFiddle Demo

rollover one image in a DIV to get another to show up in a separate DIV

Im building a website where there is a photo of a girl and to the side are a group of glasses that you can try on over her face. Right now I have it so that when you rollover one of the glasses in the group they are highlighted, but I also want a pair to show up over the photo in separate DIV when you click it.
This is what i have for just the rollover:
function rollOver()
{
document.getElementById("helm").src ="images/helmOver.jpg";
}
function rollOut()
{
document.getElementById("helm").src ="images/helmStatic.jpg";
}
</script>
<div id="framestyle">
<img class="frame" src="images/helmStatic.jpg" id="helm" border="0" width="71" height="40" onmouseover="rollOver()" onmouseout="rollOut()"/>
</div>
This is the div and image I want to show up over the photo when a pair are clicked:
<div id="glasses">
<img src="images/faceGlasses.png">
</div>
***How do i get the image in the second div to only show up when the image/rollover is clicked in the first div?
I would add
function clickedGlasses(glassesImage) {
document.getElementById("glasses_overlay").src = glassesImage;
document.getElementById("glasses_overlay").style.display = "block";
}
To your script tag at the top, and then
<img id="glasses_overlay" src="blank.png" style="display:none;">
Into your "glasses" div alongside the "faceGlasses" image (you'll need to position this over the top of the other div, either absolutely, or relatively)
And finally, change your "helm" img to
<img class="frame" src="images/helmStatic.jpg" id="helm" border="0" width="71" height="40" onmouseover="rollOver()" onmouseout="rollOut()" onclick="clickedGlasses('images/helmOverlay.png')" />

JavaScript Fade In/Out issues

I am having some difficulty trying to get my Fade In and out effect working properly. I think I am over complicating it.
I have 4 images, however only the first 2 need to be faded out and in on hover of the image (The other 2 images come into play with some other feature on the page).
My HTML is:
<div class="square">
<div class="imageHolder">
<!--Comment out and uncomment BG image to show transitions on BG images-->
<img class="one" src="image_01.jpg" />
<img class="two" src="image_02.jpg" />
<img class="three" src="image_03.jpg" />
<img class="four" src="image_04.jpg" />
</div>
</div>
Images, two, three, four are displayed none
JS:
$('.square').mouseover(function () {
$(this).find('img').each(function () {
if ($(this).attr('class') === 'two') {
$(this).fadeIn('slow');
}
if ($(this).attr('class') === 'one') {
$(this).fadeOut('slow');
}
});
});
Any help would be much appreciated.
Thanks for the responses.
I was trying to be too clever and it didn't need it. Is there a way for the the fadein and out to happen simultaneously without the use for a plugin?
Why do the each and not just selected them?
var imgs = $(this).find("img");
imgs.filter(".one").fadeOut('slow');
imgs.filter(".two").fadeIn('slow');
or
var imgs = $(this);
imgs.find(".one").fadeOut('slow');
imgs.find(".two").fadeIn('slow');
Try to do it like this:
$(".one").fadeIn("slow", function() { $(this).fadeOut("slow") });
$(".two").fadeIn("slow", function() { $(this).fadeOut("slow") });
Update:
I misread you question and thought you want both to fade in and out. To make the first one fade in and the second fade out use something like this:
$(".one").fadeIn("slow");
$(".two").fadeOut("slow");
If you have other elements with one and two classes and don't want to affect them, you can type $(".imageHolder .one") and $(".imageHolder .two") instead of $(".one") and $(".two").
If you have multiple imageHolder elements on your page, use find() function as suggested by epascarello or sushanth reddy.
You do not need a .each loop .. Just find the img inside the div and do your operations on it
Try this instead..
$('.square').mouseover(function() {
$(this).find('.two').fadeIn('slow');
$(this).find('.one').fadeOut('slow');
});​
Check FIDDLE
I think this is what you're looking for:
$('.square img')
.mouseover(function () {
$(this).fadeIn('slow');
})
.mouseout(function () {
$(this).fadeOut('slow');
});
I think you will better use jquery.hoverIntent.js. It will create a little delay time when you will move your cursor rapidly over the different images.
an example
$(document).ready(function(){
var config = {
interval: 230,
over: zoomIn,
out: zoomOut
};
$("div#clients_wrap div").hoverIntent(config);
});
zoomIn en zoomOut are functions, you could declare them with an fadein, fadeout respectively. This is just an improvement.
Basically assign a class to the group of images that need to fade in/out on hover in/out respectively
<div class="square">
<div class="imageHolder">
<!--Comment out and uncomment BG image to show transitions on BG images-->
<img class="one fadeeffect" src="image_01.jpg" />
<img class="two fadeeffect" src="image_02.jpg" />
<img class="three" src="image_03.jpg" />
<img class="four" src="image_04.jpg" />
</div>
</div>
javascript:
$('.fadeeffect')..hover(function(){
// write your code here
}

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