I'm trying to made a slideshow that slides left between images. I can get the first image to slide left, but the next image won't show up.
<script>
$(function(){
$('.fadein img:gt(0)').hide();
setInterval(function() {
$('.fadein :first-child').hide("slide", {direction: "left"}, 400)
.next('img')
.show("slide", {direction: "left"}, 400 ).end()
.appendTo('.fadein');
}, 7000);
});
</script>
HTML:
<div class="fadein">
<img style="width:600px; height:360px" src="images/fourthslide.jpg">
<img style="width:600px; height:360px" src="images/secondslide.jpg">
<img style="width:600px; height:360px" src="images/thirdslide.jpg">
<img style="width:600px; height:360px" src="images/firstslide.jpg">
</div>
CSS:
.fadein {
position:relative;
width:100%;
height:430px;
margin-left:auto;
margin-right:auto;
margin-bottom:20px;
}
.fadein img {
position:absolute;
vertical-align: middle;
top:10px;
left:0px;
border-top:1px solid navy;
}
It appears that this is an issue with the jQuery UI. The .show() method does not work as expected with the slide effect if your target element is inline. To demonstrate, I've put together a JSFiddle:
Using slide: http://jsfiddle.net/VLqjf/78/
Using swing: http://jsfiddle.net/VLqjf/79/
You will notice that if the effect is set to swing, the slideshow works as expected (case 2). However, when it is set to slide something gets messed up (case 1).
The problem is that the jQuery slide effect wraps your element in a *div.ui_effects_wrapper* that is styled position:relative. This breaks the line of images. The basic solution seems to be that you must not position your images inline.
The work-around I tested was to force the *div.ui_effects_wrapper* to be inline, using .ui-effects-wrapper{display: inline-block !important;}. Additionally, I animate the width of this div to zero as while it's being slid to the left - this effectively pushes it out of the way so the next img can slide in place. The work-around is in this JSFiddle: http://jsfiddle.net/VLqjf/105/.
Related
HTML:
<html>
<head>
<script type="text/javascript" src="jscript/jquery.js"></script>
</head>
<body>
<div class="slider_b">
<img src="img/frame_wood_back_460_380.jpg">
<img src="img/01_french_train_480_360.jpg">
<img src="img/05_cherries_480_360.jpg">
<img src="img/06_wheat_480_360.jpg">
<img src="img/10_renault_480_360.jpg">
</div>
<div id="button"><img src="img/06_wheat_480_360.jpg" width="48px" height="auto"></div>
<script>
setInterval("switchit()", 3000);
$('.slider_b img:gt(0)').hide();
function switchit() {
$('.slider_b img:first-child').fadeOut(0).next('img').fadeIn(2000).end().appendTo('.slider_b');
}
</script>
</body>
</html>
CSS:
.slider_box img{
position:relative;
top:0px;
left: 0px; }
.slider_box {
position:absolute;
width: 480px;
height: 360px; }
#button {
position: relative;
top: 10px;
left: 500px;}
The slideshow works - I just could not figure out how to switch the slideshow to one of the images by clicking a thumbnail button (id=button) - the slideshow should continue then in the regular circle order.
You could add a data attribute to your <img> elements, and append the button with the first child <img>element to carry over that data attribute. e.g:
<img src="" data-slide="1">
And for the append
var thumbnail = $("div.slider_b").find("img:first");
$("#button > img").replaceWith(thumbnail);
Once this is done, make it so that
("#button").on(click, function() {
var moveTo = $(this).find("img").data(slide);
var item = $("div.slider_b").find("[data-slide='" + moveTo + "']"
$("div.slider_b).prepend(item);
}
I'm not 100% right with the jQuery, but I believe I'm on the right lines. A bit more exploration down this route will get you to where you need to be.
Try making a relation between the slide and its thumbnail to fetch the respective slide, e.g. by using attibutes pairs, like data-slide="slide1" for the thumbnail and id="slide1" for the actual slide
on thumbnail click, adjust the current slide to the respective one and continue auto animation from this point
Point one is just one solution, it's the creativity part ;) You could come up with something else, like e.g. using thumbnails and slides indexes, etc. Good luck.
I have created a customized menu. See here. On click of this link I have a shadowbox popping up which has a long list of items. Now I want to have a "back to top" anchor link which takes me back to the top of the menu list.
I've set your lightbox with the #box id.
Html
<div id="box">
...
<!-- long content there -->
To Top
</div>
CSS (setting the width of elements)
#box {
position:relative;
width:200px;
height:250px;
overflow:auto;
}
#box #toTop {
position:absolute;
display:none;
left:150px;
top:10px;
}
jQuery
$('#box').bind('scroll', function(e) {
if ($(this).scrollTop() > 100) {
$('#toTop').fadeIn();
$('#toTop').css({'top' : $(this).scrollTop() + 100});
} else {
$('#toTop').fadeOut();
}
});
$(document).on('click', '#toTop', function(e) {
e.preventDefault();
//$('#box').scrollTop(0); //just go to top
$('#box').animate({scrollTop : 0},'slow'); //animate
});
Fiddle
Pretty easy with:
document.querySelector("iframe").contentWindow.scrollTo(0,0);
Now put a button on the page and call that on click. Oh, and omit the height:100% on your body of the iframe, this way you get rid of the second scrollbar.
You can try this out by just pasting the line above and executing it in the console of your browser with your webpage.
I'm trying to clone #main then put my ajax result there (hidden), after doing so I will make it scroll horizontally to the left hiding the current one then display the clone.
HTML:
<div class="container">
<div id="main">
<p>Click here to start</p>
</div>
</div>
CSS:
#main{
width:460px;
min-height:200px;
background:#3F9FD9;
margin:0 auto;
}
.container {
position:relative;
}
Javascript:
$('#main').click(function(){
//clone.html(data)
var clone = $(this).clone().html('<p>Ajax loaded content</p>').css(
{position:'absolute',right:'0','margin-right':'-460px',top:0}
).attr('class','love').insertAfter($(this));
$(this).css({position:'relative'});
var width = $(window).width()-$(this).outerWidth()/2;
$('#main').animate({'left':'-'+width},4000);
});
but i'm stuck on the idea on how to make both #main animate to the left and position the second div at the center?
Fiddle
EDIT: Now i'm only stuck on how to animate the clone.
I sort of took a different approach to your question, is this kind of what you are looking for?
http://jsfiddle.net/3s7Fw/5/show
I thought, rather than do some animating ourselves, why not let jQuery's hide function do it for us? This could definitely be made to work better, but it communicates the thought.
JavaScript
$('.container').on('click', '.loaded-content', function(){
$this = $(this);
//clone.html(data)
var clone = $this.clone().html('<p>Ajax loaded content</p>').attr("id", '');
$this.after(clone);
$this.hide('slow');
});
HTML
<div class="container">
<div id="main" class="loaded-content">
<p>Click here to start</p>
</div>
</div>
CSS
#main, .loaded-content{
width:460px;
min-height:200px;
background:#3F9FD9;
margin:0 auto;
float: left;
}
.container {
position:relative;
width: 920px;
}
If this is not the desired functionality, then you might be interested in a slider. There are a number of good slider plugins already out there that you can use. The difficult part would probably be adding a addNewSlide function to your chosen slider, assuming it didn't already have one.
I searched for this but didn't find an solution that totally fixed my problem.
I got 2 divs that are over each other. Where div #2 isn't shown (display:none).
Now what I want is that if I hover over div #1, div #2 slides down (open) at his current position.
Then div #2 should stay open when people are hovering over div #2, when they leave the hover status of div #2 for more then 5 seconds div #2 slides up again.
I made a fiddle to illustrate my div positions.
Using jQuery to keep the code simpler. One way to do what you want is to pair a global variable with a setTimeout function. The timeout checks if the mouse is still out of the div after five seconds, and if so, slides it up and out of sight.
$('.button').click(function() {
$('.showme').slideDown();
});
$('.showme').mouseout(function() {
window.isoverdiv = false;
setTimeout(function() {
if (!window.isoverdiv) {
$('.showme').slideUp();
}
}, 5000);
});
$('.showme').mouseover(function() {
window.isoverdiv = true;
});
http://jsfiddle.net/mblase75/TxnDd/2/
I moved div #2 into div #1 and this allowed me to do this with only css
http://jsfiddle.net/57Shn/
CSS
.button {width:100px; height:50px; position:fixed; background-color:blue; margin-top:30px;}
.button:hover .showme {display:block}
.showme {width:100px; height:200px; position:fixed; background-color:red; display:none; margin-top:30px;}
HTML
<div class="button">
touch me
<div class="showme">show me</div>
</div>
CSS-only solution: (doesn't slide)
<div class="outer">
<div class="one">Hover</div>
<div class="two">Hello World!</div>
</div>
CSS:
.two { display: none; }
.outer:hover .two { display: block; }
JS solution:
$(function() {
$('.two').hide();
$('.outer').hover(function() { $('.two').stop().slideDown(); });
$('.outer').mouseout(function() { $('.two').stop().slideUp(); });
});
I have a link and when user hover mouse over it, it should display a box (div) under the link. The box should overlay whatever is under it. How can I do it using css or javascript?
You have an absolutely positioned div that is hidden, and a child of the link. Then, when you hover over the link, you should unhide the div. I can't provide full CSS, and I haven't tested this, but that should get you started. You'll have to play around with the positioning and sizes.
Somewhere<div class="desc">This is hidden.</div>
a.special { position:relative; }
a.special div.desc { background-color:white; display:none; position:absolute; z-index:100; }
a.special:hover div.desc { display:block; }
This would be the pure-CSS way.
I have created a sample here. You can modify from there to suit your needs.
<div class="hover">Hover here</div>
<div class="overlay" style="visibility:hidden">
<img src="http://www.google.com/images/logos/ps_logo2.png" alt="google" />
</div>
$(document).ready(function()
{
$("div.hover").mouseover(function ()
{
$(this).css('cursor', 'pointer');
$("div.overlay").css('visibility','visible');
});
$("div.hover").mouseout(function ()
{
$(this).css('cursor', 'default');
$("div.overlay").css('visibility','hidden');
});
});
$("#id").mouseover(function(){
$("a[rel='#petrol']").overlay().load();
});
$("#id").mouseout(function(){
$("a[rel='#petrol']").overlay().close();
});