Multiple color background HTML - html

I'm trying to create a multiple color background to implement this:
And right now I managed to do this:
What I did:
Desired Background:
I'm trying to do it using gradients, but it seems that it's not possible to combine two gradients to do that. (It's possible to do other things, but not this).
Is there a way to implement this backgorund?
Thanks!

Try this (adjust the percentage and colors as your needs):
.yourdiv{
background: #ffffff;
background: -moz-linear-gradient(top, #ffffff 0%, #ffffff 70%, #f1f1f1 70%, #f1f1f1 100%);
background: -webkit-linear-gradient(top, #ffffff 0%,#ffffff 70%,#f1f1f1 70%,#f1f1f1 100%);
background: linear-gradient(to bottom, #ffffff 0%,#ffffff 70%,#f1f1f1 70%,#f1f1f1 100%);
filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#ffffff', endColorstr='#f1f1f1',GradientType=0 );
}
DEMO HERE

i am guessing u need the darker grey section in the desired output to be shown? if so i would suggest to divide it into sections and give individual background.
if u can post some code. i would be happy to help.

Okay, not sure if this is exactly what you want but this is how I'd do psd to css/html. See screen shot below.
Also a WORKING DEMO HERE
Just wrap the whole card in a div and apply a left border would do the trick.
border-left-width: 8px;
border-left-color: rgba(10, 255, 80, 0.75);
border-radius: 5px;
You may remove the box shadow if you don't want, just feel move active with it.

At the end, I managed to do it with this:
This for the GREEN part:
.assignment-item {
padding: 5px 5px 0px 10px !important;
margin:15px auto;
border-radius: 8px;
background: linear-gradient(to right, #4f8b2b 0%,#4f8b2b 2%,#ffffff 2%,#ffffff 100%, transparent) !important;
}
This for the GREY part:
.assignment-item:before{
position:absolute;
z-index:-1;
bottom:0;
left:2%;
width:100%;
height:25%;
content:"";
background-color:#f2f2f2;
}
Here is the result:

Related

How to put a div between two different backgrounds

I want to put a div between two different backgrounds.
It looks something like below image:
As you can see, the div placed in between two background, white and blue.
How can I achieve this?
You can use position:absolute and a negative margin-top on the div you want in the middle
Check this
You can use below code by applying to that div. In this code you can use two different colors to background that you want.
background: linear-gradient(bottom, #FFFFFF 50%, #008ED3 50%);
background: -o-linear-gradient(bottom, #FFFFFF 50%, #008ED3 50%);
background: -moz-linear-gradient(bottom, #FFFFFF 50%, #008ED3 50%);
background: -webkit-linear-gradient(bottom, #FFFFFF 50%, #008ED3 50%);
background: -ms-linear-gradient(bottom, #FFFFFF 50%, #008ED3 50%);
In above code do not remove any single line, because each line is for browser compatibility, each line code is for different browsers.
You can position your white div so that it stacks on top of the divs with blue and white background.
https://jsfiddle.net/z6ohochn/
#blue {
background: blue;
width:100%;
height:60px;
}
#white {
width:80%;a
height:50px;
background:white;
z-index:2;
position:relative;
top:-80px;
left:10%;
border-radius:5px;
}
#gray {
background: gray;
width:100%;
height:60px;
}

Bootstrap 3 sign glyphicon with white circle background

Problem
Is there any way to make a Bootstrap 3.2 sign glyph to have a white background? It is being displayed on a coloured background. I've got an example on bootply but it has a white trim that is annoying.
CSS
.glyph-white-background {
background-color:#FFFFFF;
border-radius: 50%;
}
I had a play with the bootply and there may well be better ways of doing this but for now I sorted it by placing an inner span inside the glyphicon element and positioning it so that its border does not overlap the parents.
<div class="header">
<span class="glyphicon glyphicon-exclamation-sign glyph-background">
<span class="inner"></span>
</span>
</div>
The CSS positions the inner to provide the red background for the icon only.
.header {
background-color:#3AA3CB;
font-size: x-large;
}
.glyph-background {
position:relative;
border-radius:50%;
color:#fff;
z-index:2;
}
.inner {
position:absolute;
top:2px;
left:2px;
right:2px;
bottom:2px;
border-radius:50%;
background-color:red;
z-index:-1;
}
Bootply
I followed #Duroth 's advice and it works just fine.
HTML
<span class="not-available-icon"><i class="fa fa-exclamation"></i> </span>
CSS
.not-available-icon {
background-color: #9D5A5B;
display: inline-block;
height: 25px;
width: 25px;
color: white;
border-radius: 50%;
font-size: 16px;
padding-left: 10px;
}
JS Fiddle Here
You can use any color in :
.glyph-white-background {
background-color: red;//Say red
border-radius: 50%;
}
You should specify white color for class .glyph-red
.glyph-red {
color: white;
}
See the screenshot:
http://www.bootply.com/IRTWifeP2u
I consider this absolute overkill for what you're trying to accomplish, but at least it seems to work.
Using a gradient editor, I generated an image with a radial gradient that drops from 100% to 0% opacity at around 67% / 68%, making the image fully transparent just before it hits the edge of the icon.
The following CSS should work for just about every circle icon:
.glyph-white-background {
background: url(data:image/svg+xml;base64,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);
background: -moz-radial-gradient(center, ellipse cover, rgba(255,255,255,1) 0%, rgba(255,255,255,1) 67%, rgba(255,255,255,0) 68%, rgba(255,255,255,0) 100%);
background: -webkit-gradient(radial, center center, 0px, center center, 100%, color-stop(0%,rgba(255,255,255,1)), color-stop(67%,rgba(255,255,255,1)), color-stop(68%,rgba(255,255,255,0)), color-stop(100%,rgba(255,255,255,0)));
background: -webkit-radial-gradient(center, ellipse cover, rgba(255,255,255,1) 0%,rgba(255,255,255,1) 67%,rgba(255,255,255,0) 68%,rgba(255,255,255,0) 100%);
background: -o-radial-gradient(center, ellipse cover, rgba(255,255,255,1) 0%,rgba(255,255,255,1) 67%,rgba(255,255,255,0) 68%,rgba(255,255,255,0) 100%);
background: -ms-radial-gradient(center, ellipse cover, rgba(255,255,255,1) 0%,rgba(255,255,255,1) 67%,rgba(255,255,255,0) 68%,rgba(255,255,255,0) 100%);
background: radial-gradient(ellipse at center, rgba(255,255,255,1) 0%,rgba(255,255,255,1) 67%,rgba(255,255,255,0) 68%,rgba(255,255,255,0) 100%);
}
Ofcourse, do mind your cross-browser compatibility.

Two different gradients in one html button

What I am trying to do is I want to keep the top 50% of the html button to have a gradient say from #FFF to #BBB and the bottom 50% should remain in one color lets say #111. I can't figure out a way to do it, any help would be largely appreciated.
The code of my button is:
<button class="Button1" type="submit">Submit</button>
The css:
.Button1 {
background-image: linear-gradient(to bottom, #fff 0%, #bbb 50%, #111 50%);
}
This should do the trick in latest browsers. It's up to you to make it cross-browser compatible. (I personally like the Photoshop-esque interface of http://www.colorzilla.com/gradient-editor/)
Here is a sample from Bootstrap that should help you out with button gradients. This also covers most modern browsers.
.btn-info {
color: #ffffff;
text-shadow: 0 -1px 0 rgba(0,0,0,0.25);
background-color: #49afcd;
background-image: -moz-linear-gradient(top,#5bc0de,#2f96b4);
background-image: -webkit-gradient(linear,0 0,0 100%,from(#5bc0de),to(#2f96b4));
background-image: -webkit-linear-gradient(top,#5bc0de,#2f96b4);
background-image: -o-linear-gradient(top,#5bc0de,#2f96b4);
background-image: linear-gradient(to bottom,#5bc0de,#2f96b4);
background-repeat: repeat-x;
border-color: #2f96b4 #2f96b4 #1f6377;
border-color: rgba(0,0,0,0.1) rgba(0,0,0,0.1) rgba(0,0,0,0.25);
filter: progid:DXImageTransform.Microsoft.gradient(startColorstr='#ff5bc0de',endColorstr='#ff2f96b4',GradientType=0);
filter: progid:DXImageTransform.Microsoft.gradient(enabled=false);
}
Hope that helps.(these are sort of teal, so you'll have to change that part)

Can I create gradient line effect with css or html5?

I don't want to use an image for this. I want to create a line which from transparent towards solid with css. Can I? with css3 or html5 as like this;
Like this: http://codepen.io/richbradshaw/pen/uexaG
.blurred-line {
height:30px;
width:600px;
margin:0 auto;
-moz-background-image: linear-gradient(to right, transparent 0%, black 100%);
background-image: linear-gradient(to right, transparent 0%, black 100%);
border-radius:15px;
-webkit-filter:blur(1px);
}
Which renders like:
Despite what most people seem to think, that gradient syntax is the real syntax, and works in Firefox 10+, Chrome 26+, IE10+ and Safari 6 (or 7?)+.
Including all the ancient gradient stuff is a waste of time, unless you are planning to support browsers that don't exist (e.g. Chrome 10, Firefox 3.6).
I suggest you to use a horizontal linear gradient with border radius, something like:
border-radius:50px;
background:linear-gradient(to right, rgba(0,0,0,0) 0%, rgba(0,0,0,1) 100%);
See this jsfiddle or the snippet below for more details.
.rounded {
height:50px;
width:80%;
border-radius:50px;
background: -webkit-gradient(linear, 100% 0, 0 0, from(rgba(0,0,0,0)), to(rgba(0,0,0,1)));
background: -webkit-linear-gradient(to right, rgba(0,0,0,0) 0%, rgba(0,0,0,1) 100%);
background: -moz-linear-gradient(to right, rgba(0,0,0,0) 0%, rgba(0,0,0,1) 100%);
background: -o-linear-gradient(to right, rgba(0,0,0,0) 0%, rgba(0,0,0,1) 100%);
background: linear-gradient(to right, rgba(0,0,0,0) 0%, rgba(0,0,0,1) 100%);
}
<div class="rounded"></div>
There is a gradient generator that i like a lot since it gives crossBrowser solution called "Ultimate CSS Gradient Generator".
Use rgba format
/* webkit example */
background-image: -webkit-gradient(
linear, left top, left bottom, from(rgba(50,50,50,0.8)),
to(rgba(80,80,80,0.2)), color-stop(.5,#333333)
)
An example: http://nicolahibbert.com/css3-alpha-transparent-gradients/
Duplicate: CSS3 Transparency + Gradient
This tool might be helpful too: http://www.colorzilla.com/gradient-editor/
As you said "from transparent towards solid", this should be what you want:
HTML
<div class="container">
<div class="gradient">
</div>
</div>
CSS
div.container {
width: 500px;
height: 30px;
/*background-color: #791;*/ /*uncomment this property to see the transparency effect*/
padding: 10px;
}
div.gradient {
width: 100%;
height: 100%;
background-image: linear-gradient(to right,rgba(0,0,0,0),rgba(0,0,0,1)); /* use vendor specific property if the standard one does not work */
border-radius: 25px;
}
You can do this with CSS3, and if you want it to look just like the image you provided, you can use some transparency and border-radius. I always found this link helpful:

CSS button with slanted bottom right corner

I'm trying to achieve a rectangle button with a transparent mask over the bottom right, as per this fiddle:
http://jsfiddle.net/c6gUX/
body {
background: #fff;
padding: 5em;
}
.button {
color: #FFFFFF;
font-family: 'RalewayRegular', Arial,sans-serif;
font-size: 1em;
padding: 0.5em 1.2em 0.5em 0.8em;
position: relative;
text-decoration: none;
}
.button:hover {
background: linear-gradient(to bottom, #FFA13E 0px, #E56204 100%) repeat scroll 0 0 transparent;
color: #FFFFFF;
}
.button:after {
background: url(http://i.imgur.com/8Vvw1Sw.png) no-repeat scroll 0 0 transparent;
bottom: -12px;
content: " ";
height: 38px;
position: absolute;
right: -12px;
width: 36px;
z-index: 99;
}
.orange-grad {
background: #ffa13e; /* Old browsers */
/* IE9 SVG, needs conditional override of 'filter' to 'none' */
background: url(data:image/svg+xml;base64,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);
background: -moz-linear-gradient(top, #ffa13e 0%, #ff7805 100%, #ff7805 100%); /* FF3.6+ */
background: -webkit-gradient(linear, left top, left bottom, color-stop(0%,#ffa13e), color-stop(100%,#ff7805), color-stop(100%,#ff7805)); /* Chrome,Safari4+ */
background: -webkit-linear-gradient(top, #ffa13e 0%,#ff7805 100%,#ff7805 100%); /* Chrome10+,Safari5.1+ */
background: -o-linear-gradient(top, #ffa13e 0%,#ff7805 100%,#ff7805 100%); /* Opera 11.10+ */
background: -ms-linear-gradient(top, #ffa13e 0%,#ff7805 100%,#ff7805 100%); /* IE10+ */
background: linear-gradient(to bottom, #ffa13e 0%,#ff7805 100%,#ff7805 100%); /* W3C */
filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#ffa13e', endColorstr='#ff7805',GradientType=0 ); /* IE6-8 */
}
As you can see, i've achieved it with a quite hacky way. I've untested this cross-browser and suspect IE to mess it up without a shim for :after, etc.
How can I achieve this cross-browser modern? How can I make it work when overlayed onto an image so it's effectively transparent? (See below image)
From PSD:
My fiddle on an image:
Logically I know that IE6/7 will need an image and i'm kinda OK with that.
Word length. Sometimes the button doesn't have 'Read more', so need a 100% width solution.
Edit
I am thinking of using a Sprite. (http://i.imgur.com/z0UYpTX.png)
This is tricky - particularly with your combination of a gradient with the beveled corner. The closest I could get is this fiddle, which makes use of CSS gradients to achieve the effect you're after, based on Lea Verou's awesome article.
The relevant CSS is:
.button {
background: #ffa13e; /* fallback */
background: -webkit-linear-gradient(135deg, transparent 10px, #ff7805 10px, #ffa13e 100%);
background: -moz-linear-gradient(135deg, transparent 10px, #ff7805 10px, #ffa13e 100%);
background: -o-linear-gradient(135deg, transparent 10px, #ff7805 10px, #ffa13e 100%);
background: linear-gradient(315deg, transparent 10px, #ff7805 10px, #ffa13e 100%);
}
That's the only way to get the background image to show through (that I know of). The drawback is that the gradient is not your linear one from top to bottom, but at an angle. I don't think it's possible to combine multiple gradients to match your visual precisely. But it is pretty close:
I'd suggest that this is the perfect candidate for progressive enhancement - older IE will get a solid background colour but that's perfectly acceptable (i.e. I wouldn't personally bother trying for an image fallback).
Note: there have been numerous changes between the Working Draft and the Candidate Recommendation for the gradient syntax, one of which is the angle of the gradient:
From the IE Blog. Hence the unprefixed version requires a different deg value.
Make use of it..... I think this will help You.. use it with ur image.. this will solve cross-browser issue...
http://jsfiddle.net/Praveen16oct90/2bZAW/1095/
div {
width:200px;
height:80px;
background: red;
top:150px;left:100px;
position: relative;
}
div:before {
content: '';
position: absolute;
top: 40px; right: 0;
border-right: 40px solid white;
border-top: 40px solid red;
width: 20;
}