I have a css arrow positioned at the size of my div.
Here is a jsbin that shows it in action.
How can I position the arrow half way down the div no matter what height the div is?
When you use position: absolute you can center things like this:
position: absolute;
top: 50%;
transform: translateY(-50%);
top: 50% assigns 50% of the parent's height to top
transform: translateY(-50%) moves the element up by 50% of the element's height.
This method works regardless of the element's height or the parent's height.
You can also use it to center things horizontally:
position: absolute;
left: 50%;
transform: translateX(-50%);
or both vertically and horizontally:
position: absolute;
top: 50%;
left: 50%;
transform: translate(-50%, -50%);
http://jsbin.com/lixisidezu/1/edit?html,css,output
I'd do it by changing the way you calculate top, so in this case using "calc" and by subtracting border-width.
top: calc(50% - 10px)
Here is a working jsbin.
You don't actually need CSS3 for this; CSS2 suffices and without too much hackishness: http://jsbin.com/civijuvofo/1/edit?html,css,output - the essence is to do e.g. top: 50%; margin-top: -10px;.
In particular, you can position the white+gray triangles half-way down box with top: 50%;. Then they're too low - they start at the 50% midpoint, but they descend down to 50% + 2*borderwidth. You could use css3 calc to compensate, or a css3 transform, but the most compatible hack is simply to apply a negative top margin equal to the border width.
That means your "arrow" fan-out then looks as follows in CSS:
.container:before, .container:after {
content: '';
display: block;
position: absolute;
top: 50%;
right: 100%;
width: 0;
height: 0;
border-color: transparent;
border-style: solid;
}
.container:before {
margin-top: -11px;
border-width: 11px;
border-right-color: #dedede;
}
.container:after {
margin-top: -10px;
border-width: 10px;
border-right-color: #fff;
}
This 'll work cross-browser in pretty much any browser released in the past 10-15 years. I'm not sure about IE7 :-).
Related
I am trying to set the left to be half of the width but I am unsure how to do that.
I figured you could use Calc() but I don't know how I would get the width property (set dynamically with slick.js)
.qpStar {
position: relative;
left: ???;
top: 2px;
}
I am trying to get a span to align on the left of the div using the left property. It starts at the center of the div
Obviously, left: 50% takes 50% of the offset parent.
But you might set transform: translateX(50%) instead of left.
Try This ....
.qpStar {
position: relative;
left: 50%;
top: 2px;
}
Another Way...
.qpStar {
position: relative;
transform: translateX(50%);
top: 2px;
}
I need to render a frame (ie, a picture frame around a picture) by rotating an image:
0 degrees for the left
90 for the top
... and so on.
As far as I can tell there isn't a border-image-left|right|top|bottom property, although this would work too - please correct me if I'm wrong.
It seems I'll need to use the border-image property. Does anyone know if possible to rotate the image depending on the side?
I guess the other messy options would include
Creating four div's around the image
Manually generating a frame border image (this won't really work as 1. we've got over 300 images, and 2. the frames need to be used on images with different aspect ratios... )
Edit: 'depending on the side' = 0 degrees for left, 90 degrees for top, 180 degrees for right, 240 for bottom... See image below for an example.
Left hand border image
Partially forced, but wrapping it in a div and span and playing with pseudo elements and transforms seemed to work.
The image is wrapped in an .img-container div and a span, and the ::before and ::after elements are absolute positioned around the image.
Here's the markup:
<div class="img-container">
<span>
<img src="https://unsplash.it/300/300?image=200">
</span>
</div>
And the styling:
.img-container, span, img{
display: block;
position: relative;
}
/* Image border general */
.img-container::before,
.img-container::after,
span::before, span::after{
content: "";
position: absolute;
left: -30px;
top: 50%;
width: 30px;
height: 100%;
transform: translateY(-50%);
background-image: url("https://i.stack.imgur.com/0UI1w.png");
background-size: 100% 100%;
z-index: 2;
}
/* Specific to the right border */
.img-container::after{
left: auto;
right: -30px;
top: 50%;
transform: translateY(-50%) rotate(180deg);
}
/* Top and bottom border general */
span::before,
span::after{
top: 0;
left: 50%;
transform: translate(-50%, -50%) rotate(90deg);
}
/* Just the bottom */
span::after {
top: 100%;
left: 50%;
transform: translate(-50%, -50%) rotate(270deg);
}
May be a little much
Here's a fiddle
Hello is it possible create a triangle with inverted rounded corner in the center of a rectangle, like in many landing page. Something like the below image:
I found something similar here but without inverted rounded corner
CSS Inverted Triangle image overlay
Yes, it is possible to achieve this effect by using two pseudo-elements. We need to position one of the pseudo-elements with respect to the left of the container while other is positioned with respect to right of the container. Then by adding a transform: skew() on them in opposite directions and assigning a border-radius to the required sides we can get the required output.
div {
position: relative;
height: 50px;
width: 100%;
padding-top: 50px;
background: blue;
background-clip: content-box;
/* make sure blue background doesn't appear behind triangle */
overflow: hidden;
color: white;
}
div:before,
div:after {
position: absolute;
content: '';
top: 0;
width: calc(50% + 10px);
/* don't change */
height: 50px;
/* must be equal to padding-top */
background: blue;
}
div:before {
left: 0;
transform: skew(45deg);
transform-origin: right bottom;
border-top-right-radius: 12px;
}
div:after {
right: 0;
transform: skew(-45deg);
transform-origin: left bottom;
border-top-left-radius: 12px;
}
<div class='shape'>This is a shape.</div>
I tried to make a shape using div and put an image inside. I want the image to maintain its default shape (rectangle or square) without skewing, but when I put image inside, the image skewed with the div. For the div shape I am using transform: skewY(-10deg);
.intro {
width: 180px;
height: 400px;
/* border-radius:50%;*/
cursor: pointer;
position: relative;
background: #fff;
transform: skewY(-10deg);
margin: 35px 35px 35px 0px;
}
.intro img {
width: 100%;
height: 100%;
}
<div class="intro">
<img src="http://lorempixel.com/180/400/sports">
</div>
You are trying to accomplish this: distort the shape of the outer object but keep the inner shape the same. The only way to do that is to transform the inner shape by the negative of the outer shape transform (aka, if your skewY(10deg) on the outer shape, do skewY(-10deg) on the inner), then hiding the overflow.
See this snippet:
.intro {
width: 180px;
height: 400px;
cursor: pointer;
position: relative;
background: #fff;
/* I added the -webkit- prefix as I'm using Safari 8 and
* it wouldn't show up otherwise. Might want to prefix that! */
-webkit-transform: skewY(-10deg);
transform: skewY(-10deg);
margin: 35px 35px 35px 0px;
overflow: hidden;
}
.intro img {
width: 100%;
height: 100%;
-webkit-transform: skewY(10deg);
transform: skewY(10deg);
}
<div class="intro">
<img src="http://lorempixel.com/180/400/sports">
</div>
An annoying sideeffect of this is that your contents will seem cut off. The only way to solve that is to make the inner shape larger than the outer shape an potentially padding the inside. For your image, I'd suggest:
.intro {
position: relative;
}
.intro img {
/* Use min width and heights higher than 100%
* (you might need to experiment here as it depends
* on the angle you chose for your skew) to fill
* the outer shape completely. */
min-width: 110%;
min-height: 110%;
/* Position the element absolute and 50%
* from the top and left */
position: absolute;
left: 50%;
top: 50%;
/* Now add a transform to it to move it with
* half of its width and height, therefore centering it. */
-webkit-transform: skewY(10deg) translate(-50%, -50%);
transform: skewY(10deg) translate(-50%, -50%);
}
Now you could also do width: 110%; height: 110%; left: -5%; top: -5%; and it would accomplish similar results. Play around with it.
Update
As per #vals suggestion, it might be a lot simpeler to just use the scale transform instead of all the positioning mumbo jumbo. Its always the simplest solution thats easiest to overlook:
.intro img {
-webkit-transform: skewY(10deg) scale(1.2, 1.2);
transform: skewY(10deg) scale(1.2, 1.2);
}
I want to create effect of page corner cliping like in turn.js, I know that I need two divs the outside one need to have positive rotation and inside one need the same amount but negative, and both need translate, but I don't know how to calculate the right values.
How can I do this for each corner?
Here is my try.
Here is my try, for a cut-off of 20px:
.page-wrapper {
position: absolute;
overflow: hidden;
width: 400px;
height: 300px;
top: 50px;
left: 50px;
right: auto;
z-index: 12;
background-color: blue;
}
.outter-wrapper {
position: absolute;
overflow: hidden;
-webkit-transform-origin: 100% 50%;
-webkit-transform: translateX(-20px) rotate(45deg);
right: 0px;
bottom: -100%;
width: 200%;
height: 200%;
}
.inner-wrapper {
-webkit-transform-origin: 100% 100%;
-webkit-transform: rotate(-45deg) translateX(20px);
background-color: yellow;
width: 50%;
height: 50%;
right: 0px;
position: absolute;
top: 0px;
}
You need to make the outter wrapper bigger than the inner wrapper; if not you are clipping in places that you didn't intended. I have done it 20%%; this way the math is easier.
Also, you need to adjust it carefully so that you still know the coordinates of the transform origin.
And you don't really need to move it in x and y, it's enough to mevo it horizontally.
demo