I want to create a legend using html, css which contains change of color gradient from green to yellow to red. I have tried using linear gradient property of css. However, what I got so far is given below:
#color_range {
height: 280px;
width: 40px;
background: linear-gradient(to top, #DAECB8 0%, #E33127 100%);
}
<div id="color_range"></div>
My code for color gradient
I need a figure like this:
How can I make a legend like above?
Simply change: background: linear-gradient(to top, #DAECB8 0%, #E33127 100%);
To: background: linear-gradient(red,yellow,green);
You can also change it to: linear-gradient(to top, green,yellow,red); but I don't think that to top is necessary
#color_range {
height: 280px;
width: 40px;
background: linear-gradient(red,yellow,green);
}
<div id="color_range"></div>
To understand how linear-gradient works in CSS please read: CSS Gradients
Also take a look at this page that can be helpful when using CSS gradients: https://www.colorzilla.com/gradient-editor/
You can try this.
#color_range{
height:280px;
width:40px
background:linear-gradient(red,yellow,green);
}
You can also use the color codes for these colors .
I want to apply the same gradient to the triangle (class="triangle-right") as the rectangle (class="fillblue"). I have seen some other examples but they are not working for me. Combining both shapes and using a single class would be awesome too!
JS FIDDLE HERE!
CSS:
.fillblue {
background: rgb(208,228,247); /* Old browsers */
background: -moz-linear-gradient(top, rgba(208,228,247,1) 0%, rgba(115,177,231,1) 24%, rgba(10,119,213,1) 50%, rgba(83,159,225,1) 79%, rgba(135,188,234,1) 100%); /* FF3.6+ */
background: -webkit-gradient(linear, left top, left bottom, color-stop(0%,rgba(208,228,247,1)), color-stop(24%,rgba(115,177,231,1)), color-stop(50%,rgba(10,119,213,1)), color-stop(79%,rgba(83,159,225,1)), color-stop(100%,rgba(135,188,234,1))); /* Chrome,Safari4+ */
background: -webkit-linear-gradient(top, rgba(208,228,247,1) 0%,rgba(115,177,231,1) 24%,rgba(10,119,213,1) 50%,rgba(83,159,225,1) 79%,rgba(135,188,234,1) 100%); /* Chrome10+,Safari5.1+ */
background: -o-linear-gradient(top, rgba(208,228,247,1) 0%,rgba(115,177,231,1) 24%,rgba(10,119,213,1) 50%,rgba(83,159,225,1) 79%,rgba(135,188,234,1) 100%); /* Opera 11.10+ */
background: -ms-linear-gradient(top, rgba(208,228,247,1) 0%,rgba(115,177,231,1) 24%,rgba(10,119,213,1) 50%,rgba(83,159,225,1) 79%,rgba(135,188,234,1) 100%); /* IE10+ */
background: linear-gradient(to bottom, rgba(208,228,247,1) 0%,rgba(115,177,231,1) 24%,rgba(10,119,213,1) 50%,rgba(83,159,225,1) 79%,rgba(135,188,234,1) 100%); /* W3C */
filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#d0e4f7',
endColorstr='#87bcea',GradientType=0 ); /* IE6-9 */
height: 40px;
width: 100px;
display: inline-block;
float: left;
color: white;
text-align: center;
line-height: 40px;
font-weight: bold;
}
.triangle-right {
width: 0;
height: 0;
border-top: 20px solid transparent;
border-left: 40px solid lightblue;
border-bottom: 20px solid transparent;
float: left;
}
HTML:
<div class="fillblue">Step 1</div><div class="triangle-right"></div>
Part 1: Giving the triangle a gradient
The easiest way to achieve this would be to invert your triangle. and extend the length of the element with the gradient.
JSFiddle demo.
Inverting the triangle
Rather than giving the border-left on the triangle a solid colour, you want to give the top and bototm borders the colour (in this case we want to match the background colour, so lets make these white as that's the JSFiddle background colour):
.triangle-right {
...
border-top: 20px solid white;
border-left: 40px solid transparent;
border-bottom: 20px solid white;
}
If you're unsure what this achieves, here is an example of the triangle when the top and bottom borders are set to red instead of white:
Increasing the width of your gradient element
As your triangle is 40px wide, we need to increase the width of our gradient element by 40px. For this I've used padding to ensure the text remains in the same place:
.fillblue {
...
padding-right: 40px;
}
With the same red triangle we used above, this is what it now looks like:
Positioning the inverted triangle on top of our gradient element
Now we simply need to set a negative margin on our inverted triangle to make it appear on top of our gradient element:
.triangle-right {
...
margin-left: -40px;
}
Finally, using the red triangle again, our finished result looks like this:
Part 2: Combining both shapes into one element
To do this we can make use of the :after pseudo-element.
JSFiddle demo.
First off, lets modify our HTML:
<div class="fillblue">Step 1</div>
Now lets give our .fillblue element relative positioning. We do this so that we can absolutely position our triangle in the next step:
.fillblue {
...
position: relative;
}
Now we modify our previous .triangle-right styling to use this :after pseudo-element instead:
.fillblue:after {
width: 0;
height: 0;
border-top: 20px solid white;
border-left: 40px solid transparent;
border-bottom: 20px solid white;
}
Finally we give it the new properties to position it correctly and actually make it display:
.fillblue:after {
...
content: '';
position: absolute;
top: 0;
right: 0;
}
I wanted to suggest using border-image: linear-gradient(...); but then I looked up https://developer.mozilla.org/en-US/docs/Web/CSS/border-top and saw that it's not possible to apply a border-image to just 1 of the borders, and then make the other borders transparent. There's also no border-left-image, so that won't work either. Since border-image is a relatively new addition to CSS (it's part of CSS3), it's not integrated in CSS as well as the other border styles. That's why doing this with borders is not possible. (It looks like this (simple webkit-only demo) if you do try to add a border-image, and then try to override it with transparent borders - it doesn't work)
Assuming you want to keep using borders to create your triangle, I would say this is not possible.
The only way you could make it work then is by changing the div to a square that's got a diagonal gradient, and is rotated 45 degrees via CSS transforms. That would end up being something like this:
.triangle-right {
display:inline-block;
background: -webkit-gradient(linear, left top, right bottom, color-stop(0%,rgba(208,228,247,1)), color-stop(24%,rgba(115,177,231,1)), color-stop(50%,rgba(10,119,213,1)), color-stop(79%,rgba(83,159,225,1)), color-stop(100%,rgba(135,188,234,1))); /* Chrome,Safari4+ */
/* etc. */
width:28px; /* ~ sqrt(2*40^2)/2 */
height:28px;
-webkit-transform: rotate(45deg);
/* etc. */
margin-top:6px;
margin-left:-14px;
}
Demo
Keep in mind that that is probably not the best solution, since it'd rely purely on transforms, which are not supported in every browser, and there are no good fallbacks for it. It does have one advantage over James Donnely's solution, which is that it keeps its soft borders instead of becoming jagged.
It does have other significant downsides though, namely that you're relying on fixing its position with transform and margin. It is possible other browsers don't handle this exactly the same as Chrome does, and therefore show your triangle differently. They should all show it the same way, but there's always a chance some browser decides to do things slightly differently.
Explanation of the code: The /* etc. */ stands for the other browser prefixes, the width and height are 28px because that's the height of the rotated square, its diagonal length (sqrt(width^2 + height^2)). This is also the reason the margin-left needs to be -14px (half of this diagonal length): it needs to move 14 pixels to the left, so that its corner is moved over the .fillblue element.
As was asked below in the comments, it is also possible to scale the triangle to be wider (or slimmer). This can be done by simply changing the transformation to scale(2, 1) rotate(45deg) so that it applies the stretching and rotating in the right order. A demo of this can be found at http://jsfiddle.net/x61Lyar0/2/.
PS: If you want your arrow to be less pointy, you can apply border-radius: 0 2px 0 0; (or border-top-right-radius: 2px) to smooth it out just a little bit.
Basically I'm trying to simulate Photoshop's image overlay thing using images and CSS for a menu.
There are 2 versions of the menu background image: one is the normal state (pink), and one the active state (blue). The entire menu is wrapped in a DIV with the normal (pink) image as background.
How can I make it so each active menu link uses the corresponding slice of the blue image?
Like this:
My code so far
Do you think this is possible with CSS?
CSS Only solution for modern browsers:
ul {
background-color:#ff00ff;
background-image: -moz-radial-gradient(50% 50%, ellipse closest-side, #ffffff 0%,#ff00ff 110%);
background-image: -webkit-radial-gradient(50% 50%, ellipse closest-side, #ffffff 0%,#ff00ff 110%);
background-image: -o-radial-gradient(50% 50%, ellipse closest-side, #ffffff 0%,#ff00ff 110%);
background-image: -ms-radial-gradient(50% 50%, ellipse closest-side, #ffffff 0%,#ff00ff 110%);
background-image: radial-gradient(50% 50%, ellipse closest-side, #ffffff 0%,#ff00ff 110%);
height:50px;
width:400px;
margin:0;
padding:0;
border-radius:25px;
overflow:hidden;
}
li {
width:100px;
height:50px;
float:left;
}
li:hover {
background-color:rgba(0,0,255,0.2);
}
Click to see a working demo: http://jsfiddle.net/AlienWebguy/ZLg4B/
If you need to support older browsers and can't use css3, there is a number of ways to do this. One of them:
You can cut out the blue image of the entire thing (you can actually make it wider)
then
li.active {
background: url('path/to/yourImage.png') no-repeat -50px 0;
/* 50px or however wide that rounded tip is */
}
li.active.first {
background-position: left top;
}
li.active.last {
background-position: right top;
}
/* you'll need to add 'active', 'first' and 'last' classes accordingly. */
Are you ever going to have links at the rounded parts? If not, you could just take a pixel-wide slice of the blue image and set that to the :hover state background with repeat-x.
There are definitely other ways to do this but this is the most straightforward IMHO.
Edit: After seeing your fiddle, perhaps this isn't the case. I would consider using JavaScript to calculate appropriate x-offsets for each link, and using a slice of the overlay image in that way. Or you could just make the first link a "special case" and use a generic different-color background for the rest of the links.
i have html code
<pre>
line 1
line 2
line 3
</pre>
how can i put some css style to the "lines" inside <pre>, without adding other wrapper into it?
what i mean is something like
pre lines{ color: red}
i'm having difficulties on finding the css selector for that. Thanks in advance.
You can use this little CSS3 trick, with gradients. This will create automatically, without extra spans, a "zebra" effect:
background: #555;
background-image: -webkit-linear-gradient(#555 50%, #505050 50%);
background-image: -moz-linear-gradient(#555 50%, #505050 50%);
background-image: -ms-linear-gradient(#555 50%, #505050 50%);
background-image: -o-linear-gradient(#555 50%, #505050 50%);
background-image: linear-gradient(#555 50%, #505050 50%);
background-position: 0 0;
background-repeat: repeat;
background-size: 4.5em 4.5em;
Try different CSS "line-height" so that the text appears correctly.
see: http://www.dte.web.id/2012/03/css-only-zebra-striped-pre-tag.html#.UUoV6lugkoM
If you want add color to all lines in pre, just add
pre {color: red;}
But if you want to add color to some lines, you need extra markup:
<pre>
<span>Line1</span>
line2
<span>Line3</span>
</pre>
pre span {color: red;}
I have a general rule which gives all DIVs a background image.
I have one div (with id='a') which I don't want it to have the background image.
What css rule do I have to give it?
Try:
div#a {
background-image:none
}
div#a {
background-image: none;
}
div#a {
background-image: none !important;
}
Although the "!important" might not be necessary, because "div#a" has a higher specificity than just "div".
div#a {
background-image: url('../images/spacer.png');
background-image: none !important;
}
I use a transparent spacer image in addition to the rule to remove the background image because IE6 seems to ignore the background-image: none even though it is marked !important.
Since in css3 one might set multiple background images setting "none" will only create a new layer and hide nothing.
http://www.css3.info/preview/multiple-backgrounds/
http://www.w3.org/TR/css3-background/#backgrounds
I have not found a solution yet...
When background-image: none !important; have no effect.
You can use:
background-size: 0 !important;
background-image: -webkit-gradient(linear, left top, left bottom, color-stop(0, #fff), color-stop(0.5, #fff));
background-image: -webkit-linear-gradient(center top, #fff 0%, #fff 50%);
background-image: -moz-linear-gradient(center top, #fff 0%, #fff 50%);
filter: progid:DXImageTransform.Microsoft.gradient(startColorstr='#ffffff', endColorstr='#ffffff', GradientType=0);
background-image: linear-gradient(to bottom, #fff 0%, #fff 50%);
for older browsers.. if you have defined css in some framewokrk.css like select2.css in IE9 background-image: -webkit-gradient etc. and you want it via another .css rewrite with "background-image: none !important" not works. I used same color to color gradient like page background color.
If your div rule is just div {...}, then #a {...} will be sufficient. If it is more complicated, you need a "more specific" selector, as defined by the CSS specification on specificity. (#a being more specific than div is just single aspect in the algorithm.)
HTML :
<div id="a" class="mydiv"></div>
CSS:
div#a {
background-image:none;
}
Another Way:
div:not(#a) {
//all rules goes here
//add image here
//div with id a not effected by these rules
}
Multiple (not pseudo)
div:not(#a):not(#b):not(#c) {
//all rules goes here
//add image here
//div with ids not effected with these rules
}
Doesn't this work:
.clear-background{
background-image: none;
}
Might have problems on older browsers...
Replace the rule you have with the following:
div:not(#a) { // add your bg image here //}