I need to one image overlap an another. But the second image have background color and I need the first image between the second and second's background-color. It is possible? Already tried to made a new "div class" instead of style="background-color". Now i am stuck with this:
.mainRunner {
position: relative;
}
.firstimage {
position: relative;
z-index: 2;
}
.secondimage {
position: relative;
z-index: 3;
top: -75px;
}
.background {
position: relative;
z-index: 1
}
<div class="firstimage" style="max-width: 1170px;"><img src="" alt="" title="" style="width: 100%;" max-width="1168" height="399" caption="false" /></div>
<div class="background" style="background-color: #f2e5df;">
<div class="secondimage">
<img src="" alt="" title="" />
</div></div>
You can't give certain properties of an element different z-index values. However for certain elements like a div you can use ::before and ::after pseudo elements. And you can set a z-index on those, effectively creating three layers. More information here.
In this case you can create a div with the middle img inside. Then add a ::before and ::after to that div. Giving one a background color and a z-index of -1. And the other a background image and a z-index of 1.
In the example below I also added some margin and a border around the inital div so you can better see what is going on.
.image {
margin: 20px 0 0 20px;
position: relative;
border: 3px solid coral;
width: 200px;
height: 300px;
}
.image::before,
.image::after {
content: '';
display: block;
width: 200px;
height: 300px;
position: absolute;
}
.image::before {
z-index: -1;
background: cornflowerblue;
top: 20px;
left: 20px;
}
.image::after {
z-index: 1;
background: url("https://www.fillmurray.com/200/300");
top: -20px;
left: -20px;
}
<div class="image"><img src="https://www.fillmurray.com/200/300" /></div>
If I understand right what you're trying to achieve, you probably should be placing the images within background div and placing the second image with position: absolute:
<style>
.mainRunner {
position: relative;
}
.firstimage {
position: relative;
z-index: 2;
}
.secondimage {
position: absolute;
z-index: 3;
top: 20px; /* use top and left values to place the image exactly where you want it over the first image */
left: 20px
}
.background {
position: relative;
z-index: 1;
background-color: #f2e5df;
}
</style>
<div class="mainRunner">
<div class="background">
<img src="image1.png" class="firstimage" />
<img src="image2.png" class="secondimage " />
</div>
</div>
It sets the background color as the back-most element, then on top of it the secondimage and the firstimage.
Thank everyone for their ideas. In the end the solution was simple. In the style was the double definition of second image. And the first of them was just partly commented. So my first post working right like this:
.secondimage img{
max-width: 100%;
height: auto;
position: relative;
top: -75px;
margin: 5px;
margin-bottom: 10px;
}
Now just need to find out how to close this question...
Thank you :)
The answer is simply no... there is no way to address a z-index to specifically a background of an element, z-index and all the other CSS properties work on the entire element, not on only its background.
You're going to have to find another way to do this, have you thought of using a div with not content, and the same size of the image, and then just setting a background color to that specific div?
I have a situation where, in normal CSS circumstances, a fixed div would be positioned exactly where it is specified (top:0px, left:0px).
This does not seem to be respected if I have a parent that has a translate3d transform. Am I not seeing something? I have tried other webkit-transform like style and transform origin options but had no luck.
I have attached a JSFiddle with an example where I would have expected the yellow box be at the top corner of the page rather than inside of the container element.
You can find below a simplified version of the fiddle:
#outer {
position:relative;
-webkit-transform:translate3d(0px, 20px , 0px);
height: 300px;
border: 1px solid #5511FF;
padding: 10px;
background: rgba(100,180,250, .8);
width: 80%;
}
#middle{
position:relative;
border: 1px dotted #445511;
height: 300px;
padding: 5px;
background: rgba(250,10,255, .6);
}
#inner {
position: fixed;
top: 0px;
box-shadow: 3px 3px 3px #333;
height: 20px;
left: 0px;
background: rgba(200,180,80, .8);
margin: 5px;
padding: 5px;
}
<div id="container">
Blue: Outer, <br>
Purple: Middle<br>
Yellow: Inner<br>
<div id="outer">
<div id="middle">
<div id="inner">
Inner block
</div>
</div>
</div>
</div>
How can I make translate3d work with fixed-positioned children?
This is because the transform creates a new local coordinate system, as per W3C spec:
In the HTML namespace, any value other than none for the transform results in the creation of both a stacking context and a containing block. The object acts as a containing block for fixed positioned descendants.
This means that fixed positioning becomes fixed to the transformed element, rather than the viewport.
There's not currently a work-around that I'm aware of.
It is also documented on Eric Meyer's article: Un-fixing Fixed Elements with CSS Transforms.
As Bradoergo suggested, just get the window scrollTop and add it to the absolute position top like:
function fix_scroll() {
var s = $(window).scrollTop();
var fixedTitle = $('#fixedContainer');
fixedTitle.css('position','absolute');
fixedTitle.css('top',s + 'px');
}fix_scroll();
$(window).on('scroll',fix_scroll);
This worked for me anyway.
I had a flickering on my fixed top nav when items in the page were using transform, the following applied to my top nav resolved the jumping/flickering issue:
#fixedTopNav {
position: fixed;
top: 0;
transform: translateZ(0);
-webkit-transform: translateZ(0);
}
Thanks to this answer on SO
In Firefox and Safari you can use position: sticky; instead of position: fixed; but it will not work in other browsers. For that you need javascript.
In my opinion, the best method to deal with this is to apply the same translate, but break children that need to be fixed out of their parent (translated) element; and then apply the translate to a div inside the position: fixed wrapper.
The results look something like this (in your case):
<div style='position:relative; border: 1px solid #5511FF;
-webkit-transform:translate3d(0px, 20px , 0px);
height: 100px; width: 200px;'>
</div>
<div style='position: fixed; top: 0px;
box-shadow: 3px 3px 3px #333;
height: 20px; left: 0px;'>
<div style='-webkit-transform:translate3d(0px, 20px, 0px);'>
Inner block
</div>
</div>
JSFiddle: https://jsfiddle.net/hju4nws1/
While this may not be ideal for some use cases, typically if you're fixing a div you probably could care less about what element is its parent/where it falls in the inheritance tree in your DOM, and seems to solve most of the headache - while still allowing both translate and position: fixed to live in (relative) harmony.
I ran across the same problem. The only difference is that my element with 'position: fixed' had its 'top' and 'left' style properties set from JS. So I was able to apply a fix:
var oRect = oElement.getBoundingClientRect();
oRect object will contain real (relative to view port) top and left coordinates. So you can adjust your actual oElement.style.top and oElement.style.left properties.
I have an off canvas sidebar that uses -webkit-transform: translate3d. This was preventing me from placing a fixed footer on the page. I resolved the issue by targeting a class on the html page that is added to the tag on initialization of the sidebar and then writing a css :not qualifier to state "-webkit-transform: none;" to the html tag when that class is not present on the html tag. Hope this helps someone out there with this same issue!
Try to apply opposite transform to the child element:
<div style='position:relative; border: 1px solid #5511FF;
-webkit-transform:translate3d(0px, 20px , 0px);
height: 100px; width: 200px;'>
<div style='position: fixed; top: 0px;
-webkit-transform:translate3d(-100%, 0px , 0px);
box-shadow: 3px 3px 3px #333;
height: 20px; left: 0px;'>
Inner block
</div>
</div>
Add a dynamic class while the element transforms.$('#elementId').addClass('transformed').
Then go on to declare in css,
.translat3d(#x, #y, #z) {
-webkit-transform: translate3d(#X, #y, #z);
transform: translate3d(#x, #y, #z);
//All other subsidaries as -moz-transform, -o-transform and -ms-transform
}
then
#elementId {
-webkit-transform: none;
transform: none;
}
then
.transformed {
#elementId {
.translate3d(0px, 20px, 0px);
}
}
Now position: fixed when provided with a top and z-index property values on a child element just work fine and stay fixed until the parent element transforms. When the transformation is reverted the child element pops as fixed again. This should easen the situation if you are actually using a navigation sidebar that toggles open and closes upon a click, and you have a tab-set which should stay sticky as you scroll down the page.
One way to deal with this is to apply the same transform to the fixed element:
<br>
<div style='position:relative; border: 1px solid #5511FF;
-webkit-transform:translate3d(0px, 20px , 0px);
height: 100px; width: 200px;'>
<div style='position: fixed; top: 0px;
-webkit-transform:translate3d(0px, 20px , 0px);
box-shadow: 3px 3px 3px #333;
height: 20px; left: 0px;'>
Inner block
</div>
</div>
Could someone have a look at my code. what it's suppose to do is animate the img tags using fadeIn and fadeOut but it only fades out the first img and doesn't fade in the second img. I think my css could be wrong and that's why the second image isn't showing Im not getting any errors
its an image on top of another image
jQuery
$(document).ready(function() {
$('.social-media a').on('mouseenter', function(e) {
$(this).find("img:nth-child(2)").fadeIn();
$(this).find("img:nth-child(1)").fadeOut()
});
})
HTML
<div class="social-media">
<a title="Share On Twitter" href="#">
<img alt="" src="images/icon_twitter.png" />
<img class="test" alt="" src="images/icon_twitter_active.png" />
</a>
</div>
CSS
.social-media {
padding-top: 20px;
width: 166px;
margin: 0 auto 10px auto;
}
.social-media a {
position: relative;
width: 55px;
height: 51px;
}
.social-media a img:nth-child(1) {
opacity: 1;
}
.social-media a img:nth-child(2) {
position: absolute;
left: 0; top: -33px;
opacity: 0;
z-index: 2;
}
Instead of hiding the second <img> element with zero opacity, you should use display: none instead:
.social-media a img:nth-child(2) {
position: absolute;
left: 0; top: -33px;
display: none;
z-index: 2;
}
http://jsfiddle.net/8vH4E/
However, I would strongly recommend using a simple CSS image sprite to achieve this effect, which doesn't require JS.
Update: Since OP asked if it is possible to do with CSS, I have modified the Fiddle to exclude the use of JS and simply rely on the use of CSS and pseudo-elements: http://jsfiddle.net/8vH4E/2/
.social-media a {
display: block;
position: relative;
width: 100px;
height: 100px;
background-image: url(http://placehold.it/200x200);
background-size: cover;
}
.social-media a::before {
background-image: url(http://placehold.it/200x200/4a7298/eeeeee);
background-size: cover;
content: '';
display: block;
opacity: 0;
pointer-events: none;
position: absolute;
top: 0;
left: 0;
bottom: 0;
right: 0;
-webkit-transition: all .25s ease-in-out;
transition: all .25s ease-in-out;
}
.social-media a:hover::before {
opacity: 1;
}
My strategy is rather simple:
Use background images instead. For sizing, I have used cover but you are free to use any sizing (absolute pixel/point sizes, relative percentage sizes or dynamically-computed sizes like cover, contain)
For the hover state, use an absolutely-positioned pseudo element that covers the entire <a> (by positioning it absolutely and with zero offset from all four directions). We don't need pointer events on the pseudo element, so we set it to pointer-events: none
When the <a> element is hovered on (targeted with the :hover selector), we toggle the opacity of the pseudo-element from 0 to 1. We declare the transition property on the pseudo-element to allow for smooth, browser-computed and JS-agnostic transition.
the sprite is good but does not give smooth fading animation (think that was the main reason, KDM, wasn't it?).
So let's fix existing code:
as the fadeOut() turns the element to the display: none; state, as the fadeIn() starts working when the element is display: none;. So let's turn the 2nd image in display: none; first;
We can omit the opacity at all for both images (relying on 1.0 as default); $.fadeIn/Out() use the opacity from the CSS as the start/end point of the animation. Of course you can set the opacity explicitly for each image if it's designed in such way;
display: inlibe-block; for the <a> is a good point because it contains inline elements which possibly can disappear (display: none;); that causes the the whole <a> disappearing and the mouseleave event firing with unexpected UI bugs.
Enjoy http://jsfiddle.net/8vH4E/1/ and thanks to Terry for the fiddle :)
In this fiddle http://jsfiddle.net/36Fmh/36/ I'm overlaying an image with text :
The image contains two states : a non-hover state and hover state :
How can the css be updated so that just the hover state is displayed ?
I can use jQuery to update the css to the required state on hover but im not sure what the css should be. I think I need to change the background-position attribute ?
Fiddle & code :
http://jsfiddle.net/36Fmh/36/
.qp_divSelect {
cursor: pointer;
width: 960px;
height: 240px;
background: url('http://i.imgur.com/83LeFIa.png') no - repeat;
background - position: 0px - 25px;
display: inline - block;
}
<div id="testId">
<span class="qp_divSelect" style="color: #408800;" href="javascript:void(0)" onclick="alert('here')">
<span style="position: absolute;font-weight: bold;color: white;padding-left: 22px;padding-top: 10px;font-size: 13px;top: -0px;padding-left: 22px;">Add</span>
</span>
</div>
You have to change background-position on pseudoclass :hover.
.qp_divSelect:hover {
background-position: 0 0;
}
Here my fiddle: http://jsfiddle.net/36Fmh/37/ (I had to adjust the width and height of the div).
Edit
I think I misunderstood your question. Is a possible solution for you to apply the sprite for the inner span?
E.g.: http://jsfiddle.net/36Fmh/38/
Use :hover pseudo-selector.
.qp_divSelect:hover {
cursor: pointer;
width: 960px;
height: 240px;
background: url('http://i.imgur.com/hoveres-background.png') no-repeat;
background-position: 0px -25px;
display: inline-block;
}
Here is no jQuery needs.
AS shown in image I have a [wrapper] div which has background image inside this Image I want to place another div but as the Screen size changes the background image has different dimensions and thus the position of second div must change.
I have tried jquery to get width and height of the background image but it gives out 0,0.
What should I do.
jsfiddle code jsfiddle[dot]net/AFvak/
To my knowledge, there is no facility for querying for that kind of information about a background image. The only solutions I've seen seem to involve just loading in the image by some other means (e.g. with an img tag) and then querying that for the information.
See: How do I get background image size in jQuery?
If the center div should always be centered with a fix height and width then you could try this:
<div class="wrapper">
<div class="inside"></div>
</div>
Styles:
.wrapper {
width: 600px;
height: 500px;
margin: 40px auto 0;
position: relative;
border: 1px solid black;
background: url(image_here.jpg) no-repeat center center;
}
.inside {
top: 50%;
left: 50%;
width: 200px;
height: 100px;
margin-top: -50px; /* height/2 */
margin-left: -100px; /* width/2 */
position: absolute;
background: #000;
}
DEMO
try ..
$backWidth=$(window).width();
$backHeight=$(window).height();
As per my understanding you try to div tag should be on image with fixed position even browser will resized.
Here code:
<div id="wrapper">
<div id="test">
<img src="test.jpg" id="yourimg">
<div id="yourdiv"></div>
<div>
</div>
<style>
#test{
position:relative;
}
#yourimg{
position:absolute;
top:100px;
left:100px;
}
#yourdiv{
position:absolute;
top:120px;
left:120px;
}
</style>