horizontal and vertical lines in a square - html

Below is the image I am trying for, I managed to get a square using CSS, but I am trying for horizontal and vertical line in a square.
.hub{
width: 119px;
height: 101px;
background: #b5adad;
}
<div class="hub"></div>

There are many ways to do this and one would be to use gradients like below: (the image in question was actually a rectangle.)
The approach is very simple - we use 2 linear gradients to create two thin solid colored lines and then position the images such that they match our needs. Linear gradients are used even though it creates only a solid color because it is easier to control size and position of an image than background color.
div {
height: 100px;
width: 200px;
border: 1px solid red;
background-image: linear-gradient(to bottom, red, red), linear-gradient(to right, red, red);
background-repeat: no-repeat;
background-size: 1px 100%, 100% 1px;
background-position: 20px 0px, 0px 10px;
}
<div></div>
We can also create an output which has a fade-out or shadow effect like in the image in question:
div {
height: 100px;
width: 200px;
border: 1px solid;
background-color: gray;
background-image: linear-gradient(to bottom, black, black), linear-gradient(to right, red, transparent), linear-gradient(to right, black, black), linear-gradient(to bottom, red, transparent);
background-repeat: no-repeat;
background-size: 1px 100%, 1px 100%, 100% 1px, 100% 1px;
background-position: 20px 0px, 21px 0px, 0px 10px, 0px 11px;
box-shadow: inset 0px 0px 3px red;
}
<div></div>

Another way is to use :before and :after pseudo-elements:
.hub{
width: 119px;
height: 101px;
background: #b5adad;
position: relative;
padding: 18px 0 0 18px;
}
.hub:after, .hub:before {
content: " ";
background: black;
display: block;
position: absolute;
}
.hub:after {
width: 1px;
height: 100%;
left: 15px;
top: 0;
}
.hub:before {
width: 100%;
height: 1px;
top: 15px;
left: 0;
}
<div class="hub">Lorem ipsum dolor amet</div>

Related

How to create a responsive zig-zag border using only css. The starting and ending pattern must match

I want to create a zig-zag border in css which is responsive, i.e. the zig-zag border must adjust itself to fit perfectly according to width of the container.
I was able to create this:
But on changing the width it's output is :
I want to perfectly fit the zig-zag pattern like above image on changing the width of the container.
It would be helpful if I could also add some radius at peak points like this :
Here is the code so far
.container {
width: 664px;
}
.sub-container {
border: 2px solid black;
border-bottom: 0;
padding: 40px;
height: 200px;
}
.upper-zigzag {
background: linear-gradient(135deg, #ffffff 25%, transparent 25%) 0px 0,
linear-gradient(225deg, #ffffff 25%, transparent 25%) 0px 0;
background-size: 60px 60px;
background-color: black;
height: 32px;
background-repeat: repeat-x;
border-left: 2px solid black;
border-right: 2px solid black;
}
.lower-zigzag {
position: relative;
background:
linear-gradient(315deg, #ffffff 25%, transparent 25%) -28px -30px,
linear-gradient(45deg, #ffffff 25%, transparent 25%) -28px -30px;
background-size: 60px 60px;
background-color: transparent;
height: 30px;
background-repeat: repeat-x;
margin-top: -30px;
z-index: 1;
}
<div class="container">
<div class="sub-container"></div>
<div class="upper-zigzag"></div>
<div class="lower-zigzag"></div>
</div>
Thanks!

Create pentagon shape css :before with background image

I need to create a "div top" with a pentagon shape with a background inside.
Is this possible with CSS3? I have tried a number of things with not success. This is the code I am currently left with:
.sectionDIVHead:before {
position: absolute;
float: left;
content: " ";
background-size: 50% 80%;
background-color: inherit;
width: 40px;
height: 40px;
margin-left: calc(50% - 95px);
margin-top: -5%;
background-position: bottom;
border-left: 52px solid transparent;
border-right: 54px solid transparent;
background-image: url("../images/Icon.png");
background-repeat: no-repeat;
background: linear-gradient(to top left, #000 50%, transparent 51%)0% 100%/50.5% 70px no-repeat, linear-gradient(to top right, #000 50%, transparent 51%)100% 100%/50% 70px no-repeat;
padding: 35px;
padding-bottom: 0;
text-align: center;
}

Create transparent css border with triangle

I'm trying the find a way to have a dynamic border with a triangle. For the moment, with the basic gradient effect, this is what I did:
My current effect in action
But as you can see, the background has a gradient and we can see the border background that does not match..
How can I achieve this effect? Also, the text may vary on different screen size and with other words.
Thank you!
Using pseudo-elements and skewX is one clean way to achieve this. Check this out, I'm using a top, left & right border on the element, and then style the before as the left bottom border and the after as the right one:
body {
background-color: white;
background-image: linear-gradient(45deg, #999 25%, transparent 25%, transparent 75%, black 75%, black), linear-gradient(45deg, black 25%, transparent 25%, transparent 75%, #999 75%, #999);
background-size: 10px 10px;
background-position: 0 0, 50px 50px;
}
.dialog {
text-align: center;
color: green;
font-size: 65px;
width: 300px;
height: 120px;
background-color: transparent;
border-width: 5px 5px 0 5px;
border-color: red;
border-style: solid;
display: inline-block;
position: relative;
}
.dialog:before {
content: '';
border-top: 5px solid red;
border-right: 5px solid red;
transform-origin: left top;
transform: skewX(45deg);
position: absolute;
content: ' ';
height: 10px;
width: 46%;
background: inherit;
left: -5px;
bottom: -10px;
}
.dialog:after {
content: '';
border-top: 5px solid red;
border-left: 5px solid red;
transform-origin: left top;
transform: skewX(-45deg);
position: absolute;
content: ' ';
height: 10px;
width: 46%;
background: inherit;
right: -5px;
bottom: -10px;
}
<div class="dialog">Here I am</div>
To achieve this you can make it with a background image, for exemple http://bootsnipp.com/snippets/featured/carousel-reviews-with-rating.
As you can see he take an image and resize it to take only a triangle like this:
.sprite-i-triangle {
background-position: 0 -1298px;
height: 44px;
width: 50px;
}
Try to find an image that meets your expectations. Otherwise you have some exemples in this site. (http://bootsnipp.com)

How to remove some part of borders from the corners?

I want to remove the corners of borders like this picture.
You can use ::before and ::after pseudo elements to cover (and thus, "hide") parts of the border:
.bordery {
border: 1px solid teal;
padding: 20px;
position: relative;
}
.bordery::after,
.bordery::before {
background-color: white;
content: "";
display: block;
height: 10px;
position: absolute;
width: 10px;
}
.bordery::after {
bottom: -1px;
right: -1px;
}
.bordery::before {
top: -1px;
left: -1px;
}
<div class="bordery">This is just some sample content.</div>
You can use :before and :after pseudo elements to create this.
.el {
position: relative;
width: 200px;
height: 50px;
margin: 50px;
}
.el:after,
.el:before {
content: '';
position: absolute;
height: 90%;
width: 100%;
}
.el:before {
top: -5px;
left: -5px;
border-top: 1px solid orange;
border-left: 1px solid orange;
}
.el:after {
bottom: -5px;
right: -5px;
border-bottom: 1px solid orange;
border-right: 1px solid orange;
}
<div class="el"></div>
You can use css3 linear-gradient to draw this background to just a single <div> element and without using any pseudo elements.
div {
background-image: linear-gradient(to left, transparent 20px, orange 20px),
linear-gradient(to bottom, transparent 20px, orange 20px),
linear-gradient(to right, transparent 20px, orange 20px),
linear-gradient(to top, transparent 20px, orange 20px);
background-position: 100% 0, 100% 0, 0 100%, 0 100%;
background-size: 100% 1px, 1px 100%;
background-repeat: no-repeat;
}
div {
background-color: #eee;
background-image: linear-gradient(to left, transparent 20px, orange 20px),
linear-gradient(to bottom, transparent 20px, orange 20px),
linear-gradient(to right, transparent 20px, orange 20px),
linear-gradient(to top, transparent 20px, orange 20px);
background-position: 100% 0, 100% 0, 0 100%, 0 100%;
background-size: 100% 1px, 1px 100%;
background-repeat: no-repeat;
position: relative;
margin: 0 auto;
height: 100px;
width: 80%;
}
<div></div>
You can do it by following:
#rectangle{
width:400px;
height: 200px;
border-style: solid;
color:orange;
position: absolute;
}
#eraser1{
position: absolute;
width: 50px;
height: 50px;
background-color:white;
margin: -10px 0px 0px 374px;
}
#eraser2{
position: absolute;
width: 50px;
height: 50px;
background-color:white;
margin: 175px 0px 0px -13px;
}
Use clip-path property to clip corner
div{
width: 15em;
height: 5em;
border: 2px solid red;
clip-path: polygon(0 0, 91% 0, 100% 12%, 100% 100%, 12% 100%, 0 89%);
}
<div></div>

how to add gradient to borders

i was wondering if its possible to add gradients to border top without it affecting border right or border left which in this case are transparent. i tried adding a gradient color but it would affect border left and border right im trying to let border left and border right to be transparent
#borderone {
border-top: 33px solid #354658;
border-left: 33px solid transparent;
border-right: 33px solid transparent;
margin: 0 auto;
min-width: 1277px;
}
<div id='borderone'></div>
as you can see this is what i want it to do although i want a gradient background color instead of this solid dark blue color http://jsfiddle.net/EHELN/
See this :
http://css-tricks.com/examples/GradientBorder/
It is enough for me in my career .
For example:
#borderone:first-child:before {
content:'';
position:absolute;
width:100%;
height:4px;
background:linear-gradient(to left, #354658, #9EBBDA);
top:-33px;
left:-5;
}
For your case , you should use before & first-child pseudo-selectors CSS in the same time.
top(in pseudo selector) = -border height = -33 px
FIDDLE: http://jsfiddle.net/abdennour/EHELN/2/
You can get this efect using background for the gradient, and the 2 pseudo elements at the left and right to get the slanted corners
.test {
border-left: 33px solid transparent;
border-right: 33px solid transparent;
height: 33px;
background: linear-gradient(90deg, black, blue);
margin: 0 auto;
min-width: 42px;
background-clip: content-box;
position: relative;
}
.test:before, .test:after {
content: '';
position: absolute;
width: 33px;
height: 100%;
}
.test:before {
background: linear-gradient(45deg, transparent 50%, black 50%);
right: 100%;
}
.test:after {
background: linear-gradient(315deg, transparent 50%, blue 50%);
left: 100%;
}
demo
Looks like I missunderstood the direction. Try this to make it the other way (for webkit)
.test {
border-left: 33px solid transparent;
border-right: 33px solid transparent;
height: 33px;
background: linear-gradient(0deg, black, red);
margin: 0 auto;
min-width: 42px;
background-clip: content-box;
position: relative;
}
.test:before, .test:after {
content: '';
position: absolute;
width: 45px;
height: 45px;
bottom: 0px;
}
.test:before {
-webkit-transform: rotate(45deg);
-webkit-transform-origin: bottom right;
background: linear-gradient(-45deg, black 0, red 32px, transparent 32px);
right: 100%;
}
.test:after {
-webkit-transform: rotate(-45deg);
-webkit-transform-origin: bottom left;
background: linear-gradient(45deg, black 0, red 32px, transparent 32px);
left: 100%;
}
demo 2
if you want to draw a gradient on your border, then you could use border-image or translucide borders with a gradient in bg : DEMO
But then :
You can even drop your translucide borders and make it a padding: DEMO
#borderone {
position:relative;
padding:33px 33px 0;/* well this is just like transparent borders :) */
margin: 0 auto;
height:100px;
background:linear-gradient(
to bottom,
#354658,
#9EBBDA 33px,
transparent 33px
);
}
http://www.colorzilla.com/gradient-editor/
This is for the background and not the border, but you can likely create the same effect you are looking for by using this tool.