Consider the following attempt to rotate a paragraph 90 degrees and position it so that the corner that was initially its top-left corner (and which therefore becomes its top-right corner after the rotation) ends up located at the top-right corner of the parent block.
HTML:
<!DOCTYPE html>
<html>
<body>
<div id="outer">
<p id="text">Foo bar</p>
</div>
</body>
</html>
CSS:
#outer {
border: solid 1px red;
width:600px;
height: 600px;
position: relative;
}
#text {
transform: rotate(90deg);
position: absolute;
top: 0;
right: 0;
}
In Firefox 19.0.2 on OS X 10.6.8, it fails. This appears to be because, despite the order in which the CSS properties were given, the transformation is applied after the positioning. In other words, the browser:
places #text such that its top-right corner is located at the top-right corner of the parent block, but only then
rotates it, with the result that what is now its top-right corner is not located at the top-right corner of the parent block.
As a result, the transform-origin property isn't much use here. If, for instance, one used transform-origin: top right; then #text would need to be moved downwards by the width it had before it was rotated.
My question: is there a way to tell the browser to apply the CSS positioning properties after the rotation; and if not, then is there instead a way to move #text downwards (e.g. using top:) by the width it had before it was rotated?
NB. Ideally the solution should not require setting a fixed width: for #text, and must not require JavaScript.
You can apply more than one transform to an element, and the order does matter. This is the simplest solution: http://jsfiddle.net/aNscn/41/
#outer {
border: solid 1px red;
width:600px;
height: 600px;
position: relative;
}
#text {
background: lightBlue;
position: absolute;
top: 0;
left: 0;
right: 0;
transform: translate(100%) rotate(90deg);
transform-origin: left top;
-webkit-transform: translate(100%) rotate(90deg);
-webkit-transform-origin: left top;
}
The transform origin is the point around which a transformation is applied. For example, the transform origin of the rotate() function is the center of rotation - https://developer.mozilla.org/en-US/docs/Web/CSS/transform-origin
Rotating -90deg.
.rotate {
position:absolute;
-webkit-transform-origin: left top;
/* Safari */
-webkit-transform: rotate(-90deg) translateX(-100%);
/* Firefox */
-moz-transform: rotate(-90deg) translateX(-100%);
/* IE */
-ms-transform: rotate(-90deg) translateX(-100%);
/* Opera */
-o-transform: rotate(-90deg) translateX(-100%);
}
Solved: here
This is the code I've added:
left: 100%;
width: 100%;
-webkit-transform-origin: left top;
I've also added some prefixed transform properties so it will be cross browser
-webkit-transform:rotate(90deg);
-moz-transform:rotate(90deg);
-ms-transform:rotate(90deg);
-o-transform:rotate(90deg);
transform:rotate(90deg);
How I did it:
I've found this question and, as the name of the website says, "fiddled" with the code to obtain this behavior. I guess the solution is left: 100%; instead of right: 0;.
(the width: 100%; is there because for some reason it wasn't 100% and the text would overflow to the next line)
You may want to try using CSS3 #keyframes animation. It will allow you to rotate and reposition in any order you like. Here is a tutorial that may help: [CSS-Tricks][1]
.container {
position: relative;
width: 200px;
height: 200px;
border: 1px solid red;
}
p {
border: 1px solid blue;
position: absolute;
top: auto;
right: 0;
display: inline-block;
margin: 0;
animation: 1s rotate 1s both;
}
#keyframes rotate {
0% {
transform-origin: top left;
transform: rotate(0deg);
right:0;
}
50% {
right:0;
}
100% {
transform-origin: top left;
transform: rotate(90deg);
right: -64px;
}
}
<div class="container">
<p>some text</p>
</div>
You might want to play around with the translate option which you can apply as the second transform function after rotate and place your element at the exact position that you want to.
There is no other way I guess to tell the browser to use the position properties after the transform function is used using plain css.
See this demo - http://codepen.io/anon/pen/klImq
Place "!important" at the end of the transform line.
Related
My end goal is a draggable, resizable, Scalable, and rotatable element, just like the example on: https://daybrush.com/moveable/ only by using css width,height, and transform: rotate, translate.
Say I have a div with following css:
.rect {
background-color: red;
width: 200px;
height: 100px;
top:100px;
left:100px;
position: absolute;
transform: rotate(0deg);
}
<div class="rect"></div>
If I want to resize the div horizontally to the left, I just change the width by x pixels. If I want to change it to the right I just change the width by x pixels, and translate(-xpx, 0).
But what if I change the angle? From trying a lot of stuff, I found some of the x and y values for translate to the respective angle, however I feel like there is a more straight forward way than just guessing. E.g: For 90deg, if I want to resize to the left by x px I do translate(-x0.5px, x0.5px).
More: what if I want to change both the width & height at the same time?
P.S.: I would rather avoid using libraries, transform: scale or svg
P.P.S:Example to further demonstrate the problem, just changing the width:
.rect {
background-color: red;
width: 200px;
height: 100px;
top:100px;
left:100px;
position: absolute;
transform: rotate(45deg);
animation: expand 5s infinite
}
#keyframes expand {
from {width: 200px;}
to {width: 2000px;}
}
<div class="rect"></div>
Fixed, stretching the left side of the original rectagle (now up since rotated 90deg):
.rect {
background-color: red;
width: 200px;
height: 100px;
top:100px;
left:100px;
position: absolute;
transform: rotate(90deg);
animation: expand 3s infinite
}
#keyframes expand {
from {
width: 200px;
}
to {
width: 800px;
transform: rotate(90deg) translate(-300px, 300px);
}
}
<div class="rect"></div>
You can apply several transformations to the same object and they will be composed in the order that you specify. Move then rotate, is different than rotate then move.
.rect {
background-color: red;
width: 200px;
height: 100px;
top:100px;
left:100px;
position: absolute;
}
.t1 {
background-color: #40d04080;
/* green shaded rectangle: rotate after translation */
transform: translate(2cm, 0) rotate(30deg);
}
.t2 {
background-color: #f0404080;
transform: rotate(30deg);
}
.t3 {
background-color: #4040f080;
/* blue shaded rectangle: translate after rotation */
transform: rotate(30deg) translate(2cm, 0) ;
}
<div class="rect t1"></div>
<div class="rect t2"></div>
<div class="rect t3"></div>
I have a square and its animation code up with pure html and css. Here is the jsbin url to the code: https://jsbin.com/medupun/edit?html,output
And following is the most relevant CSS part:
.foo {
height: 100px;
width: 100px;
-webkit-transform-origin: 100px 0%;
}
.foo:hover {
-webkit-transform: rotate(-180deg);
}
The square is 100px by 100 px. I want its transform origin to be the
top right corner of itself . I am seeking to do it with just x/y-offset. So I do it with "-webkit-transform-origin: 100px 0%;".
However, I see the square does not rotate around the top right corner, rather a point close to it. And what surprises me is, if I change the origin to 130px 0%, it will work.
Can someone help me understand where is the extra 30px comes from?
You have padding on your square that adds the 30px.
padding: 1em;
Remove that to get the results you are looking for.
.foo {
position: absolute;
top: 400px;
left: 400px;
height: 100px;
width: 100px;
text-align: center;
background: #ffea61;
-webkit-transition: all 750ms ease-in-out;
-webkit-transform-origin: 100px 0;
}
.foo:hover {
-webkit-transform: rotate(-180deg);
}
https://jsbin.com/zovucefuxi/edit?html,output
I am trying to understand what is really happening “3d” world of CSS.
I made a simple example
Particularly the code which bugs me the most is:
.back {
background-color: tomato;
transform: rotateY(180deg);
z-index: 1;
}
The thing which is not clear to me is why when you hover over .inner, its background color (gold) is not visible?? If you remove the transform property from .back or if you set the rotateY to 0deg then the gold background color of the .inner is clearly visible.
Why is the transform property of .back changing the stacking order?
Logically it makes sense that children(.front and .back) should appear in front of their parent(.inner).
Also, I would like to know what really happens when you set transform-style to flat? Does that make parent and all of its children collapse into single “unit” where element with highest stacking order takes priority/visibility?
in your code :
.outer {
display: block;
width: 200px;
height: 200px;
border: 2px solid gold;
perspective: 1000px;
padding: 10px;
margin: 0 auto;
}
.inner {
position: relative;
height: 100%;
width: 100%;
transition: transform 2s linear;
transform-style: preserve-3d;
background-color: gold;
backface-visibility: visible;
transform: rotateY(50deg);
}
.sides {
position: absolute;
width: 100%;
height: 100%;
left: 0;
top: 0;
color: white;
backface-visibility: hidden;
}
.front {
background-color: blue;
transform: translateZ(20px)
}
.back {
background-color: tomato;
transform: rotateY(180deg) translateZ(10px);
}
.inner:hover {
transform: rotateY(180deg)
}
<div class="outer">
<div class="inner">
<div class="sides front">Front Side</div>
<div class="sides back">Back Side</div>
</div>
</div>
you are using
transform: rotateY(180deg) translateZ(10px);
The transforms are applied right to left, so first it goes to the front 10px. But after that, it rotates 180deg. (around the transform-origin that is constant). That makes the previous 10px go towards the back instead of to the front.
if the order is the inverse
transform: translateZ(10px) rotateY(180deg);
now the rotation is done first, and so the translation is unafected by it and goes to the front.
and No, sorry, z-index is not a substitute for 3-d transforms, if you want to use 3d transforms, translation is the only way to go ....
In your first example, z-index is useless, as can be seen easily
codepen with z-index removed
This works because you are setting
backface-visibility: hidden;
So only the face that is facing front will be visible
I am performing a CSS transform: rotate on a parent, yet would like to be able to negate this effect on some of the children - is it possible without using the reverse rotation?
Reverse rotation does work, but it affects the position of the element, and it may have a negative performance impact (?). In any case, it doesn't look like a clean solution.
I tried the "transform: none" suggestion from this question prevent children from inheriting transformation css3, yet it simply doesn't work - please see the fiddle here: http://jsfiddle.net/NPC42/XSHmJ/
May be you have to write like this:
.child {
position: absolute;
top: 30px;
left: 50px;
background-color: green;
width: 70px;
height: 50px;
-webkit-transform: rotate(-30deg);
-moz-transform: rotate(-30deg);
-o-transform: rotate(-30deg);
-ms-transform: rotate(-30deg);
transform: rotate(-30deg);
}
Check this for more http://jsfiddle.net/XSHmJ/1/
Updated:
You can use:after & :before psuedo class for this.
check this http://jsfiddle.net/XSHmJ/4/
I believe that you are going to need to fake it using a second child, the specification does not seem to allow for the behavior you would like, and I can understand why the position of a child element has to be affected by a transform to its parent.
This isn't the most elegant of solutions, but I think you're trying to do something that the specification is never going to allow. Take a look at the following fiddle for my solution:
.parent {
position: relative;
width: 200px;
height: 150px;
margin: 70px;
}
.child1 {
background-color: yellow;
width: 200px;
height: 150px;
-webkit-transform: rotate(30deg);
-moz-transform: rotate(30deg);
-o-transform: rotate(30deg);
-ms-transform: rotate(30deg);
transform: rotate(30deg);
}
.child2 {
position: absolute;
top: 30px;
left: 50px;
background-color: green;
width: 70px;
height: 50px;
}
<div class="parent">
<div class="child1"></div>
<div class="child2"></div>
</div>
If you want to apply transforming effects on a parent without affecting its children, you can simply animate a parent's pseudo-element like this:
.parent {
display: inline-block;
position: relative;
}
.parent::before {
content: "";
background: #fab;
/* positioning / sizing */
position: absolute;
left: 0;
top: 0;
/*
be aware that the parent class have to be "position: relative"
in order to get the width/height's 100% working for the parent's width/height.
*/
width: 100%;
height: 100%;
/* z-index is important to get the pseudo element to the background (behind the content of parent)! */
z-index: -1;
transition: 0.5s ease;
/* transform before hovering */
transform: rotate(30deg) scale(1.5);
}
.parent:hover::before {
/* transform after hovering */
transform: rotate(90deg) scale(1);
}
This actually worked for me. JSFiddle
I want to rotate the image which is placed in the button of scrollbar in Chrome. Now I have a CSS with this content:
::-webkit-scrollbar-button:vertical:decrement {
background-image: url(images/arrowup.png);
-webkit-transform: rotate(120deg);
-moz-transform: rotate(120deg);
background-repeat: no-repeat;
background-position: center;
background-color: #ECEEEF;
border-color: #999;
}
I wish to rotate the image without rotating its content.
Very well done and answered here – http://www.sitepoint.com/css3-transform-background-image/
#myelement:before {
content: "";
position: absolute;
width: 200%;
height: 200%;
top: -50%;
left: -50%;
z-index: -1;
background: url(background.png) 0 0 repeat;
transform: rotate(30deg);
}
Very easy method, you rotate one way, and the contents the other. Requires a square though
#element{
background : url('someImage.jpg');
}
#element:hover{
transform: rotate(-30deg);
}
#element:hover >*{
transform: rotate(30deg);
}
Update 2020, May:
Setting position: absolute and then transform: rotate(45deg) will provide a background:
div {
height: 200px;
width: 200px;
outline: 2px dashed slateBlue;
overflow: hidden;
}
div img {
position: absolute;
transform: rotate(45deg);
z-index: -1;
top: 40px;
left: 40px;
}
<div>
<img src="https://placekitten.com/120/120" />
<h1>Hello World!</h1>
</div>
Original Answer:
In my case, the image size is not so large that I cannot have a rotated copy of it. So, the image has been rotated with photoshop. An alternative to photoshop for rotating images is online tool too for rotating images. Once rotated, I'm working with the rotated-image in the background property.
div.with-background {
background-image: url(/img/rotated-image.png);
background-size: contain;
background-repeat: no-repeat;
background-position: top center;
}
Good Luck...
CSS:
.reverse {
transform: rotate(180deg);
}
.rotate {
animation-duration: .5s;
animation-iteration-count: 1;
animation-name: yoyo;
animation-timing-function: linear;
}
#keyframes yoyo {
from { transform: rotate( 0deg); }
to { transform: rotate(360deg); }
}
Javascript:
$(buttonElement).click(function () {
$(".arrow").toggleClass("reverse")
return false
})
$(buttonElement).hover(function () {
$(".arrow").addClass("rotate")
}, function() {
$(".arrow").removeClass("rotate")
})
PS: I've found this somewhere else but don't remember the source
I was looking to do this also. I have a large tile (literally an image of a tile) image which I'd like to rotate by just roughly 15 degrees and have repeated. You can imagine the size of an image which would repeat seamlessly, rendering the 'image editing program' answer useless.
My solution was give the un-rotated (just one copy :) tile image to psuedo :before element - oversize it - repeat it - set the container overflow to hidden - and rotate the generated :before element using css3 transforms. Bosh!
try making a div for the image only and then flipping it with transform: scaleY(-1); or transform: scaleX(-1);
if you want to have the navbar in front of the image you can make an overlapping div and set its opacity property to 0;
I tried all solutions but none helped, below is what was my problem and how I solved it:
Problem: we have an image for desktops with landscape orientation but To show the same image but rotated (portrait) for mobile screens.
How: I just rotated the actual image in my assets folder the way I wanted (portrait), and then just used media queries to call that image for my background for mobiles, and that's it.
(this was the easiest and quick solution I did.)
Update Dec 2021
Since the original question is
"..rotate the background image .."
The best answer looks to be here
https://stackoverflow.com/a/62135576/3446280