I'm trying to make my website's background to have gradient and a repeating pattern image. The gradient works fine as below, but if I remove background-repeat: no-repeat; the gradient doesn't stretch and repeats every few lines. I want to add another pattern image over the gradient but it needs to repeat so conflicts with the gradient.
I thought I can fix it by adding a div which contains the pattern image and stretches all over the body but it was unsuccessful. Is there any way to fix this problem?
body {
...
background-repeat: no-repeat;
background-attachment: fixed;
background-image: linear-gradient(bottom, rgb(255,255,255) 5%, rgb(171,205,139) 53%, rgb(171,205,139) 77%);
background-image: -o-linear-gradient(bottom, rgb(255,255,255) 5%, rgb(171,205,139) 53%, rgb(171,205,139) 77%);
background-image: -moz-linear-gradient(bottom, rgb(255,255,255) 5%, rgb(171,205,139) 53%, rgb(171,205,139) 77%);
background-image: -webkit-linear-gradient(bottom, rgb(255,255,255) 5%, rgb(171,205,139) 53%, rgb(171,205,139) 77%);
background-image: -ms-linear-gradient(bottom, rgb(255,255,255) 5%, rgb(171,205,139) 53%, rgb(171,205,139) 77%);
background-image: -webkit-gradient(
linear,
left bottom,
left top,
color-stop(0.05, rgb(255,255,255)),
color-stop(0.53, rgb(171,205,139)),
color-stop(0.77, rgb(171,205,139))
);}
And this is how I'm trying to repeat pattern image
background-image:url('mypattern.png');
background-repeat:repeat;
It seems like you will most likely need to create two elements one positioned absolutely over the other. There is a similar question here on stackoverflow dealing with this issue
LINK
CSS gradient above background pattern
EDIT
something like this may work
.example3 {
background-image: url(../images/plus.png), -moz-linear-gradient(top, #cbe3ba, #a6cc8b);
background-image: url(../images/plus.png), -webkit-gradient(linear, left top, left bottom, from(#cbe3ba), to(#a6cc8b));
}
LINK
http://www.heresonesolution.com/2010/09/using-a-css-gradient-and-a-background-image-on-the-same-element/
Related
I have a couple questions about Linear Gradient:
Is it possible to have an actual image rather than colour display instead of either the #000000 of #ffffff?
background-image: -webkit-linear-gradient(30deg, #000000 50%, #ffffff 50%);
Also would it be possible in the above example (which is black for 50% width then a 30 degree vertical split then white for 50%), so if the image replaced #000000, is it possible to place a border on the right hand of the image along the 30 degree divide that seperated the image and colour?
Thanks!
I am not sure I got right what do you need, but here is an example.
.gradient-image {
width:128px;
height:128px;
background:
linear-gradient(to right,
rgba(0,0,0,0) 0%,
rgba(0,0,0,0) 60%,
rgba(0,0,0,1) 61%,
rgba(0,0,0,1) 65%,
rgba(0,0,0,0.7) 66%,
rgba(0,0,0,0.7) 100%),
url(https://i.stack.imgur.com/ZXvxw.jpg?s=128&g=1);
}
<div class="gradient-image"></div>
I am having problems inserting a background image and making my header and footer sections have a gradient background. My background image has to repeat. Does the .gif picture have to be in the same folder as the css? This is the css for it.
body {
background-image: url("folder1/pic.gif");
background-repeat: repeat-x repeat-y;
}
I am also working on a linear gradient that goes from white to orange to black. This gradient would then be the background for the header and footer sections. I am able to split and provide the header and footer section a linear gradient background but it does not extend all the way to the border of my header and footer. This is the html section for the header
<header>
<div id="eg1">
<img src="images/pumpkin.gif" alt= "pumpkin" height="78" width="85">
<h1>The Halloween Store</h1>
<h3>For the little Goblin in all of us!</h3>
</div>
</header>
The css formatting the html I have as
/*gradient header*/
#eg1 {
background-image: -webkit-linear-gradient(45deg, white 0%, #ffa500 75%, #000000 100%);
}
As you can see in this example the gradient shows fine and the background image also shows.
Be aware: On some browsers the gradient may not appear because of the browser version or the syntax.
So I used all the prefixes for gradient to be as many brosers as possible are supported:
background-image: linear-gradient(45deg, white 0%, #ffa500 75%, #000000 100%);
background-image: -webkit-linear-gradient(45deg, white 0%, #ffa500 75%, #000000 100%);
background-image: -moz-linear-gradient(45deg, white 0%, #ffa500 75%, #000000 100%);
background-image: -o-linear-gradient(45deg, white 0%, #ffa500 75%, #000000 100%);
For more about browser support check this
fairly new to doing multiple gradients. I've got a linear gradient, then a radial one on top.
background-image: -webkit-gradient(
linear,
left bottom,
left top,
color-stop(0, #033968),
color-stop(1, #408BCD)
);
background-image: url('../img/background-noise.png'), -o-linear-gradient(top, #033968 0%, #408BCD 100%);
background-image: url('../img/background-noise.png'), -ms-linear-gradient(top, #033968 0%, #408BCD 100%);
background-image: url('../img/background-noise.png'), -moz-radial-gradient(center top, farthest-side, rgba(255,255,255,1) 0, rgba(255,255,255,0) 100px), -moz-linear-gradient(top, #033968 0%, #408BCD 100%);
background-image: url('../img/background-noise.png'), -webkit-radial-gradient(center top, farthest-side, rgba(255,255,255,1) 0, rgba(255,255,255,0) 100px), -webkit-linear-gradient(top, #033968 0%, #408BCD 100%);
background-image: url('../img/background-noise.png'), radial-gradient(farthest-side at center top, rgba(255,255,255,1) 0, rgba(255,255,255,0) 100px), linear-gradient(to top, #033968 0%, #408BCD 100%);
It works fine, but I'm wondering why the radial gradient's size is defined in pixels, but it act's like a %.
Check out: http://jsfiddle.net/tK5Ch/
When you adjust the browser size, the radial gradient moves around. I just want to make it fixed, like 100px x 100px
Any ideas what's happening?
Since you have two separate axes defined, the default shape is ellipse. You need to specify that it is circle:
http://jsfiddle.net/tK5Ch/3/ (Only changed the standard gradient)
radial-gradient(circle farthest-side at center top, ...
I want to make a CSS3 gradient that consists of a 1px line.
How can I do this?
I have tried the following code, but the gradient that is produced is too thick:
background-image: linear-gradient(left , rgb(255,255,255) 50%, rgb(209,209,209) 50%, rgb(255,255,255) 51%);
(see here)
How can I make the line smaller, so its only 1px wide? The percentage values seem to control the positioning of the line, but no matter how much I adjust them, I can't get it to 1px wide!
(Essentially, I am using the line to act as a 'faux columns' background [i.e. to visually separate a left and right column. (Although, to keep the jsFiddle simple, I have removed the columns)] I know there are other ways of doing columns, but this method is the best for my situation)
EDIT: Just to clarify, this is for a slightly odd use case, where the width has to be 100% and no psudeo-elements can be used.
/* Opera Mobile */
background: -o-linear-gradient(left, #d1d1d1 1px, white 1px);
/* Firefox for Android */
background: -moz-linear-gradient(left, #d1d1d1 1px, white 1px);
/* WebKit browsers */
background: -webkit-linear-gradient(left, #d1d1d1 1px, white 1px);
/* new syntax: IE10, Firefox, Opera */
background: linear-gradient(90deg, #d1d1d1 1px, white 1px);
background-position: 100% 0;
background-repeat: repeat-y;
background-size: 50%;
demo
[I used 2px instead of 1px in the demo as 1px was not visible. I only tested in Chrome though.]
You should always put the unprefixed version last. There is no need for -ms-linear-gradient. IE10 now supports the standard syntax with no prefix and IE9 doesn't support gradients at all.
If you don't care about IE8 (which you probably don't if you're using gradients) you can use calc().
background-image: linear-gradient(left, transparent 50%, rgb(255,255,255) 50%, rgb(255,255,255) calc(50% + 1px), transparent calc(50% + 1px));
This is will work with any width element, whereas just using percentages will break down on smaller and wider elements.
.style {
background-image: -o-linear-gradient(left , rgb(255,255,255) 50%, rgb(209,209,209) 50%, rgb(255,255,255) 50.5%);
background-image: -moz-linear-gradient(left , rgb(255,255,255) 50%, rgb(209,209,209) 50%, rgb(255,255,255) 50.5%);
background-image: -webkit-linear-gradient(left , rgb(255,255,255) 50%, rgb(209,209,209) 0%, rgb(255,255,255) 50.5%);
background-image: linear-gradient(left , rgb(255,255,255) 50%, rgb(209,209,209) 50%, rgb(255,255,255) 50.5%);
}
You are not dealing with pixels, you are using percentages. So 1% of your width, which must be 200 is 2px. (I think that is why this works, maybe I'm wrong.) You can use percentages decimals, so .5% == 1px.
I had use this earlier, change it according to your need. I mean change colors and angle as you want
background-image: liner-gradient(to bottom, white, white 14%,blue 1%,white 15%);
I have a which I am going to make into a button. The top half should be #ffd41a and the bottom half should be #fac915. Here is a link to the button at present. http://jsfiddle.net/WnwNW/
The problem that I'm facing is how should I deal with two background colors. Is there a way to do what I'm trying to do without the need for addition divs or spans? Can I have two background attributes within the same CSS class?
CSS3 provides a way to do this
background-image: linear-gradient(to bottom, #FFD51A 50%, #FAC815 50%);
background-image: -o-linear-gradient(bottom, #FFD51A 50%, #FAC815 50%);
background-image: -moz-linear-gradient(bottom, #FFD51A 50%, #FAC815 50%);
background-image: -webkit-linear-gradient(bottom, #FFD51A 50%, #FAC815 50%);
background-image: -ms-linear-gradient(bottom, #FFD51A 50%, #FAC815 50%);
http://jsfiddle.net/WnwNW/1/
Yes and no. You can use two background attributes. However, this is only supported in CSS3. That means that two background images will break in older browsers. That being said, you can do something like this.
background-image: url(color1.png), url(color2.png);
background-position: bottom, top;
background-repeat: no-repeat;
I'm not sure if you can specify multiple background "colors."