I'm using the following code to give the double underline on my site (https://howtogetrippedathome.com/) below my widget titles a gradient color:
.widget-title:after {
border-bottom: 6px double;
-webkit-border-image: -webkit-linear-gradient(left, #ff2828, #F27B26);
}
However, when I apply this code the underline dissapears. I have looked in other topics and this should work, but I don't know what I'm doing wrong.
Simply use multiple gradient like this
h1{
display:inline-block;
padding-bottom:5px;
background:
linear-gradient(to left, red, blue),
linear-gradient(to left, red, blue);
background-size:100% 2px;
background-position:bottom 0 left 0,bottom 5px left 0;
background-repeat:no-repeat;
}
<h1>some text</h1>
Instead of using the pseudo element(:after),
Try this directly:
.widget-title {
border-bottom: 6px double;
-webkit-border-image: -webkit-linear-gradient(left, #ff2828, #F27B26);
}
In css3 please use pseudo-element with ::after instead of :after.
And please ensure you at least an empty content for the pseudo element style like (content: "") and specify display property.
.widget-title {
width: 100px;
height: 100px;
}
.widget-title::after {
content: "";
display: block;
background: #ffba10;
border-bottom: 6px double;
}
Above code works as expected. Please refer this link for more info.
Related
I'm trying to create a button that has two colors as a border.
The two colors i need used are blue: #00a7e1, orange: #f6531d.
I would like to just use css if possible.
Thank in advance!
link to button concept
Example:
.btn
{
border: 0;
padding: 4px;
display: inline-block;
background: linear-gradient(20deg, #00a7e1 49%, #e65300 50%);
}
.bg
{
background: #349645;
padding: 8px 14px;
font: bold 24px Consolas;
}
.btn:active .bg
{
background: #0a1117;
color: #ffffff;
}
<div class="btn"><div class="bg">YOU'R TITLE</div></div>
<button class="btn"><div class="bg">YOU'R TITLE</div></div>
You may also play with gradient and background-clip (see comments in CSS)
button {
vertical-align: top;
border: 5px solid transparent;/* give extra space for gradients colors */
font-size: 2.5rem;
margin: 0.25em;
padding: 0.5em 2em;
background: linear-gradient(#333, #333),/* black turned into gradient to hold with background-clip and hide the 2 color gradient under it */
linear-gradient(/* 2 colors to draw under the borders also via background-clip*/
to bottom left,
rgb(230, 83, 0) 50%,
gray 51%,
rgb(0, 166, 224) 40%
)
no-repeat center center;
background-clip:
padding-box, /* drawn inside including padding area */
border-box;/* drawn also under borders */
background-size:
100% 100%,
110% 150%;/* must be bigger than 100% so it include also borders, else it repeats */
color: white;
box-shadow: 0 0 2px 2px black, inset 0 0 2px black;/* did you want this too ? */
}
<button>BUTTON</button> <button> TO</button> <button> PLAY</button>
If you think this is too much, you also have border-image .
Simply use border-image with a gradient:
button {
padding:20px;
border:5px solid;
border-image:linear-gradient(60deg,#00a7e1 50%,#f6531d 0) 20;
background:transparent;
}
<button>some text</button>
I have this box with a linear gradient background created as a two tone solid color. One color is 44px - the rest has another color, like this:
background: linear-gradient(to left, #365aa5 44px, #f5f5f5 0);
Works great. Now I would like to add a two-tone border to the top and bottom of this element using border image linear gradients the same way - so that the colors of the border follow the color of the background. The trick is to use linear gradients as solid colors.
I have tried something like this:
border-image: linear-gradient(right, #365aa5 44px, #000 0);
border-style: solid;
border-width: 2px 0 2px 0;
But obviousley, it's not working.
Any ideas how I could make this work?
JsFiddle here.
You need to add a number in the end of the border-image property. In your case it has no effect but it is still required. Also use to right instead of right
div {
height: 50px;
width: 80%;
padding: 4px;
box-sizing: border-box;
position: relative;
background: linear-gradient(to left, #365aa5 44px, #f5f5f5 0);
/* What I'm trying: */
border-image: linear-gradient(to right, #365aa5 44px, #f5f5f5 0) 1;
border-style: solid;
border-width: 2px 0 2px 0;
}
body {
padding: 20px;
background-color: #fff;
}
<div>Two tone solid color top and bottom border to<br> match the two tone background</div>
I took the blue color so it is easier to see.
EDIT: Also possible as vibhu suggested:
border-image: linear-gradient(to right, #365aa5 44px, #f5f5f5 0);
border-image-slice: 1;
You can add the two tone border by using the below additional code::
div::after {
content: "";
position: absolute;
height: 2px;
width: 44px;
right: 0;
background: #365aa5;
top: -2px;
}
div::before {
content: "";
position: absolute;
height: 2px;
width: 44px;
right: 0;
background: #365aa5;
bottom: -2px;}
Jsfiddle added here: https://jsfiddle.net/y2Ln2h86/
Hi i've got a input range on html5 min 0 and max 100.
But i would like to color a part for example between 70 and 100.
I don't want to use bootstrap for this.
I don't know how to do that.
You can easily do this by using a linear-gradient as background for the track. All that we need to do is create a gradient which is colored only for the width that we need (30% for your case because you need it colored only between 70-100) and then position it with respect to the track's (the track is the bar of the range input) right side. Since the styling of range inputs is still in experimental phase we have to use browser prefixed selectors (to select the track of each browser) and then apply styles to it. We also have to do some additional corrections to address browser specific problems, I've marked these with inline comments in the code.
The below code is tested and found to be working fine in Edge, IE11 and latest versions of Chrome, Firefox and Opera (all on a Windows 10 machine).
Note: This will only color the part between 70-100 of the range input differently. This doesn't have the code to make the appearance of range input the same in all browsers. I've not done that because that is out of the scope of this question.
Also, as mentioned by ssc-hrep3 in his comment, this may not be good for production implementation because these things are still in experimental stage and we've to use browser specific selectors but if you want to apply custom styling to HTML5 range inputs then there is probably no other way.
input[type=range] {
-webkit-appearance: none;
border: 1px solid black; /* just for demo */
}
input[type=range]::-webkit-slider-runnable-track {
background: linear-gradient(to left, red 30%, transparent 30%);
background-position: right top;
}
input[type=range]::-moz-range-track {
background: linear-gradient(to left, red 30%, transparent 30%);
background-position: right top;
}
input[type=range]::-ms-track {
background: linear-gradient(to left, red 30%, transparent 30%);
background-position: right top;
background-repeat: no-repeat; /* no repeat means background appears a little on the left due to width issue and hence the fix */
width: 100%; /* to fix width issue in Edge */
color: transparent; /* to avoid the intermediate stripe lines in < IE11 */
border: none; /* just do away with the track's border */
}
input[type=range]::-ms-fill-lower {
background: transparent; /* IE11 has default fill and that needs to be removed */
}
<input type="range" min="0" max="100" value="70" step="10" />
For the benefit of future readers: Just in case you need uniform styling across all major browsers then you could use the below snippet. It produces almost similar output in all of them.
input[type=range] {
-webkit-appearance: none;
}
input[type=range]::-webkit-slider-runnable-track {
background: linear-gradient(to left, red 30%, transparent 30%);
background-position: right top;
height: 10px;
box-shadow: inset 0px 0px 0px 1px black;
}
input[type=range]::-moz-range-track {
background: linear-gradient(to left, red 30%, transparent 30%);
background-position: right top;
height: 10px;
box-shadow: inset 0px 0px 0px 1px black;
}
input[type=range]::-ms-track {
background: linear-gradient(to left, red 30%, transparent 30%);
background-position: right top;
background-repeat: no-repeat; /* no repeat means background appears a little on the left due to width issue and hence the fix */
width: 100%; /* to fix width issue in Edge */
height: 10px;
color: transparent; /* to avoid the intermediate stripe lines in < IE11 */
border-color: transparent;
border-style: solid;
border-width: 10px 0px; /* dummy just to increase height, otherwise thumb gets hidden */
box-shadow: inset 0px 0px 0px 1px black;
}
input[type=range]::-ms-fill-lower {
background: transparent; /* IE11 has default fill and that needs to be removed */
}
input[type=range]::-webkit-slider-thumb {
-webkit-appearance: none;
height: 18px;
width: 18px;
margin-top: -4px;
background: sandybrown;
border: 1px solid chocolate;
border-radius: 50%;
}
input[type=range]::-moz-range-thumb {
height: 18px;
width: 18px;
background: sandybrown;
border: 1px solid chocolate;
border-radius: 50%;
}
input[type=range]::-ms-thumb {
height: 18px;
width: 18px;
margin-top: 0px; /* nullify default margin */
background: sandybrown;
border: 1px solid chocolate;
border-radius: 50%;
}
<input type="range" min="0" max="100" value="70" step="10" />
I'm trying to create a button with CSS that will sit on a semi-transparent background that has a beveled or cut edge to it. Here is the Photoshop mockup:
I'm able to do this successfully with a solid color background because I can use an pseudo element with that same background and "cover" the edge of the button, but it doesn't work with a semi-transparent background.
Here's what I've got so far, on a solid background: http://codepen.io/anon/pen/GJFpc
I'm beginning to believe this isn't possible with just CSS, but still hoping S.O. can save me once again!
I love a good css challenge so I tried a few things and this is what I could come up with:
http://jsfiddle.net/QE67v/3/
The css (unprefixed) looks like this:
a.cta {
position: relative;
float: left;
padding: 8px 10px;
text-align: center;
text-decoration: none;
text-transform: uppercase;
font-size: 15px;
font-weight: normal;
background-image: linear-gradient(top, #ffffff 0%, #e4e4e4 100%);
box-shadow: inset 0 -2px 1px 2px #fff;
line-height: 16px;
height: 16px;
z-index: 2;
}
a.cta:after {
content: '';
display: block;
position: absolute;
width: 32px;
height: 32px;
right: -16px;
top: 0;
background-image: linear-gradient(top, #ffffff 0%, #e4e4e4 100%);
box-shadow: inset -3px -2px 1px 2px #fff;
transform: skewX(-45deg);
z-index: -1;
}
There are two main differences with your code:
I use a inset box-shadow to achieve the white 'bevel'. You could
probably do this with gradients as well, but I just find the shadows
more intuitive.
In stead of making the button wider and covering the bottom left
corner with a pseudo element in the color of the background, I kept
the button in its normal width and added a pseudo element to which a
applied the skewX transformation. This allows for any background, as
you can see by the gradient I set as a background in my fiddle.
I believe this is what you where after. Feel free to ask if you need any further help/explanation.
Here's my code:
<h3>Desserts <span>+</span></h3>
And here's my style that accompanies said code:
#our_menu h3 span {
color: #eee;
display: block;
float: right;
font-size: 11.5pt;
margin-top: -25pt;
margin-right: 10pt;
text-shadow: 0px 1px 0px #240d00; }
#our_menu h3:first-child a {
border-top: 1px solid #692601;
background: -webkit-gradient(
linear,
left bottom,
left top,
color-stop(0.1, rgb(201,86,19)),
color-stop(1, rgb(159,58,0))
); }
This is being used on a mobile site that I'm building and for some reason the span is being killed by the webkit gradient. Has anyone experienced anything like this before? If I remove the gradient then the span reappears :(
Thanks!
Figured it out. The span is not part of the link and the gradient style is being applied to the link which renders over top of the span. I made the span relative with a z-index and it works fine. Good grief.