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Image positioned below to be visible on mouse hover

There are two images placed one below the other, what i want to do is when you hover a mouse over the top image only the portion of the image below should be visible not the entire image to be replaced.Is this possible using jquery ?.
I am stuck on where to start. I tried changing the div background on hover but i couldn't get anywhere near what i need.Thanks.
html
<div style="background-image: url("image1.jpg")></div>
<div style="background-image: url("image2.jpg")></div>

Try this?
$(document).ready(function() {
var $hover = $("#hover");
var $foreground = $("#foreground");
$hover.hide();
$foreground.mousemove(function(event) {
var top = event.pageY - $hover.height() / 2;
var left = event.pageX - $hover.width() / 2;
$hover.css("top", top);
$hover.css("left", left);
});
$foreground.mouseover(function() {
$hover.show();
});
$foreground.mouseleave(function() {
$hover.hide();
});
});
http://jsfiddle.net/WV8jX/685/
EDIT: Updated to hide before mouse entering
http://jsfiddle.net/WV8jX/686/
EDIT:
It doesn't really solve your problem tho if you need the background to be shown as according to your description.

Since you have tagged jQuery, use the jQuery .hover().
Example:
$("selector").hover(
function() {
show your hidden image functionality here
}
);
https://api.jquery.com/hover/

I think there is no need of using Javascript/Jquery. CSS can do the job as below
HTML
<div class="bgdemo"></div>
CSS
.bgdemo{
background-image: url("image1.jpg");
}
.bgdemo:hover{
background-image: url("image2.jpg");
}
and there is a typo in your HTML, style tag is not ending properly.
edit
Check this jsfiddle, Position the background according to your requirement.
edit
Here I found good article seems bit complicated but worth reading..

Make scrollTo make div scroll all the way?

I have a div with only horizontal overflow. With a link outside the div, I'm trying to scroll the div to a certain image (think of a horizontal gallery scrolling to the right).
I used the following javascript. It works fine in the webpage.
However, the DIV containing the gallery is larger than most images. Consequently the browser window will scroll only until the requested div comes in from the right and is now fully on screen, and not one pixel more. However, I would like the div to scroll all the way, so that the image is all the way hugging the left edge of the container.
I hope I'm making sense, I'm not terribly experienced, but I couldn't find an answer to my question online.
$(document).ready(function(){
$('#gimg1').click(function() {
$.scrollTo($('#gimg1link'), 1000);
});
$('#gimg2').click(function() {
$.scrollTo($('#gimg2link'), 1000);
});
$('#gimg3').click(function() {
$.scrollTo($('#gimg3link'), 1000);
});
$('#gimg4').click(function() {
$.scrollTo($('#gimg4link'), 1000);
});
});
<div id="gallery">
<img class="galleryimage" id="gimg1" src="lb1.jpg">
<img class="galleryimage" id="gimg2" src="lb2.jpg">
<img class="galleryimage" id="gimg3" src="lb3.jpg">
<img class="galleryimage" id="gimg4" src="lb4.jpg">
</div>
Image 1
Image 2
Image 3
Image 4

You are using the image and link selectors in your jquery in the wrong order.
$('#gimg1').click(function() {
$.scrollTo($('#gimg1link'), 1000);
});
This snippet means "when the image #gimg1 is clicked, scroll to the position of the link #gimg1link". You want it the other way round: when the link is clicked, scroll to the image.
Reversing those selectors gives you a working slider: jsFiddle
The last image will always stay on the right of the screen, because that's where the document ends and it can't scroll any further. The other images will scroll all the way to the left as long as your document width allows it.
Also, you could optimize your javascript a lot by not copy-pasting the same code but just making it more generic:
$(document).ready(function(){
$('a[id^=gimg]').click(function() { // when a link with an ID starting with "gimg" is clicked
var linkID = $(this).attr('id'); // get the whole id from this link
var imgID = linkID.replace('link', ''); // get the img ID it relates to by removing the 'link' part
$scrollTo( $('#' + imgID), 1000); // scroll to the image this link belongs to
});
});
Now it doesn't matter how many links and images you add, as long as they all use the same naming convention.

Based on this answer i adapted the code to suit your need. It uses the clicked thumbnail index to find the corresponding image left, and set scrollLeft of the viewport to this value.
$('#nav li').click(function(){
var clickedIndex = $(this).index();
var targetElement = $('#viewport ul li').eq(clickedIndex);
var elementPosition = targetElement.position();
$('#viewport').animate({scrollLeft: elementPosition.left},500);
});
Working fiddle: http://jsfiddle.net/Lqvqtwtb/

JavaScript How to make an image to change color onmouseover?

I have a button and an image and want them to change color onmouseover.
The button changes color fine:
<script>
function secondColor(x) {
x.style.color="#000000";
}
function firstColor(x) {
x.style.color="#ffaacc";
}
</script>
<input onmouseover="secondColor(this)" onmouseout="firstColor(this)" type="submit"><br>
How can I do the same thing with the image? Is there any way:
<img src="..." ......
Or do I have to have a second image to replace the first one onmouseover and this is the only way?

If you don't care that much about supporting older browsers, you could use the new CSS3 filter brightness. In chrome, you could write something like this:
var image = document.getElementById('img');
image.addEventListener('mouseover', function() {
image.setAttribute('style','-webkit-filter: brightness(1.5)');
}, false);
image.addEventListener('mouseout', function() {
image.setAttribute('style','-webkit-filter: brightness(1.0)');
}, false);
I don't recommend this approach, though. Using another picture while hovering would be a better solution.

I know that this is old, but you don't need two images. Checkout my example using one image.
You can simply change the position of the background image.
<div class="changeColor"> </div>
JavaScript
var dvChange = document.getElementsByClassName('changeColor');
dvChange[0].onmouseover = function(){
this.style.backgroundPosition = '-400px 0px';
}
dvChange[0].onmouseout = function(){
this.style.backgroundPosition = '0px 0px';
}
CSS
.changeColor{
background-image:url('http://www.upsequence.com/images/multibg.png');
width:400px;
height:400px;
background-position: 0px 0px;
}
.changeColor:hover{
background-image:url('http://www.upsequence.com/images/multibg.png');
width:400px;
height:400px;
background-position: -400px 0px;
}
You can also try changing the opacity of the images onmouseover and onmouseout.
I don't have an example for that, but its super easy to find and I am sure it has be answered already on stack exchange somewhere.
In the JSFiddle below there is Javascript and non-Javascript examples.
http://jsfiddle.net/hallmanbilly/gtf2s8ts/
Enjoy!!

I think you have to use a second image. I recently cam across the following article describing how to do image crossfading on hover using css. Crossfading Image Hover Effect

You can change image SRC on mouse over, you can load two images and use fade effects to "change" them. But better, you can use image as DIV background, make sprite and just move BG on mouse over.
Loading of two different images bring you to disappearing when hover and second image loading. Better do not use JS at all. Make sprite from two images, put it as BG of DIV and write two CSS for DIV, normal and when hover.

If you have access to JQuery use hover function. If you want to change image
$('#imageid').hover(function(){
//change image or color or opacity
$(this).attr('src', newImageSrc);
});
add this function in document ready function.

Is it possible to reveal the contents of a div from right to left

I created a fiddle
http://jsfiddle.net/gifcy/bJJ5s/5/
On DOM ready I hide the image.
I could successfully reveal the image from left to right using animate function.
Can someone show how to reveal from right to left. What additional parameters need to used.

Use jquery ui show() function
$(document).ready(function() {
$('#nature').hide();
$('#animation').click(function() {
$('#nature').show('slide', {direction: 'right'},1000);
});
});
​
fiddle here http://jsfiddle.net/bJJ5s/6/

Pretty sure this will work cross browser
#nature,
#nature img {
float:right;
}​
http://jsfiddle.net/g8zDk/

Div width expand/shrink on click

For a site I'm making for myself and a friend, I have a div container/wrapper with 2 other divs within it: one occupies the left half and has a black background and the other occupies the right with a white background. Essentially, this lets me get a split colored background. Each div holds half of a logo. Here's the page, temporarily hosted so you guys can see it.
http://djsbydesign.com/tempsite/index.htm
At any rate, I'd like to have links on the left and right hand sides of the page that, on click, cause their respective divs to expand from 50% to 100%. I have a few ideas, but am not sure entirely how to go about doing this (I'm rather new to javascript). The first would be to have the expanding div's z-index set to something higher than the non-expanding one, and then have it expand (somehow), and the other is to have the expanding div expand to 100% while the other shrinks to 0% at an equal rate.
The bottom line is, I have no idea how to go about doing this. I don't mind using mootools or jQuery, for the record.

The following seems to work:
$('#left-bg, #right-bg').click(
function(){
$(this).animate({'width': '100%'},600).siblings().animate({'width':'0'},600);
});
Albeit I'm not sure how you'd plan to bring back the the 'other' div.
JS Fiddle demo.
Edited to add a button (via jQuery) that allows both divs to be reverted to original dimensions:
$('#left-bg, #right-bg').click(
function(){
$(this).animate({'width': '100%'},600).siblings().animate({'width':'0'},600);
$('<button class="show">Show all</button>')
.appendTo('#wrapper');
});
$('.show').live('click',
function(){
$('#left-bg').animate(
{
'width': '50%'
},600);
$('#right-bg').animate(
{
'width': '50%'
},600);
$(this).remove();
});
Updated JS Fiddle.
Edited to address the question left by OP in the comments:
is there a way to have a page redirect after the animation completes?
Yep, just add the line window.location.href = "http://path.to.url.com/";
$('#left-bg, #right-bg').click(
function(){
$(this).animate({'width': '100%'},600).siblings().animate({'width':'0'},600);
$('<button class="show">Show all</button>')
.appendTo('#wrapper');
window.location.href = "http://www.google.com/" // <-- this line redirects.
});
$('.show').live('click',
function(){
$('#left-bg').animate(
{
'width': '50%'
},600);
$('#right-bg').animate(
{
'width': '50%'
},600);
$(this).remove();
});
Updated JS Fiddle.
Edited in response to bug report (in comments):
The one other bug (easy fix) is that any time you click on either of the divs, it creates a new button. So say you clicked on the left half, and it expanded and filled the page, etc., and then you clicked on it again (it being anywhere on the page now). It would attempt to add a second button.
To prevent a second button being added to the div just add an if:
$('#left-bg, #right-bg').click(
function(){
if (!$('.show').length) {
$(this).animate({'width': '100%'},600).siblings().animate({'width':'0'},600);
$('<button class="show">Show all</button>')
.appendTo('#wrapper');
window.location.href = "http://www.google.com/" // <-- this line redirects.
}
});
Which, will only append a button, or indeed animate the divs, so long as the $('.show') selector returns no matches.
However if you're also redirecting to another page by clicking the button it shouldn't be an issue anyway, since none of the jQuery on the original page will exectute/be able to access the page to which the user is redirected (unless it's a page on your own domain, and you've explicitly chosen to add the same button).

If you give absolute positions to your div's such that - 1st is positioned at top left corner and other is positioned at top right corner. And then in click event you can change the position of the other top corner of the div to be expanded.
You can use jquery to do this easily. Check jquery documentation for setting css.

Looks like you've got jQuery included, so use that! It's totes the easiest library to do simple animations with.
Here's an example click function that will slide the right background to be 100% like you said:
$('a#link').click(function(e) {
e.preventDefault();
$('#left-bg').animate({ width : '0%' }, 'slow');
$('#right-bg').animate({ width : '100%' }, 'slow');
});
Obviously to go in the other direction you'd switch the width values in the object passed to the animate functions.
If you're not familiar with the animate function, check the docs, but basically you just pass CSS rules in a key : value object to it, and it'll change the CSS values over time - animating it!
Hope this helps!