Browser Handling of CSS "Transparent" in Gradients

Browser handling of CSS “transparent” in gradients

1. Although the Color module states that transparent means the same as rgba(0, 0, 0, 0), colors behave a little differently in CSS gradients. The Image Values module states that color stops should be interpolated in a premultiplied RGBA color space. This means that browsers are supposed to preserve the RGB colors during transitions between color-stops, and that the grey semitransparent tones shouldn't be there.

2. As of the end of October 2012, only IE10 and Opera perform this interpolation correctly, such that the grey portions aren't present and that you get a pure white 0%-to-100% alpha gradient. Other browsers display the grey portions, which is incorrect.

CSS3 gradient rendering issues from transparent to white

I've encountered this as well. I'm not sure why it happens, but here's what I've used in my own projects as a workaround:

background-image: -webkit-linear-gradient(top, rgba(255,255,255,0.001) 0%, #fff 5%, #fff 100%);

Instead of giving Chrome a "transparent" value, give it something very, very close to transparent. Hope this helps!

Edit: I forgot to post a link to my own version, which renders as expected in Chrome 21 (Windows 7).

Transparent Background Image with a Gradient

Keep in mind that a CSS gradient is actually an image value, not a color value as some might expect. Therefore, it corresponds to background-image specifically, and not background-color, or the entire background shorthand.

Essentially, what you're really trying to do is layering two background images: a bitmap image over a gradient. To do this, you specify both of them in the same declaration, separating them using a comma. Specify the image first, followed by the gradient. If you specify a background color, that color will always be painted underneath the bottom-most image, which means a gradient will cover it just fine, and it will work even in the case of a fallback.

Because you're including vendor prefixes, you will need to do this once for every prefix, once for prefixless, and once for fallback (without the gradient). To avoid having to repeat the other values, use the longhand properties¹ instead of the background shorthand:

#mydiv .isawesome { 
    background-color: #B1B8BD;
    background-position: 0 0;
    background-repeat: no-repeat;

    /* Fallback */
    background-image: url('../images/sidebar_angle.png');

    /* CSS gradients */
    background-image: url('../images/sidebar_angle.png'), 
                      -moz-linear-gradient(top, #ADB2B6 0%, #ABAEB3 100%);
    background-image: url('../images/sidebar_angle.png'), 
                      -webkit-gradient(linear, left top, left bottom, color-stop(0%, #ADB2B6), color-stop(100%, #ABAEB3));
    background-image: url('../images/sidebar_angle.png'), 
                      linear-gradient(to bottom, #ADB2B6, #ABAEB3);

    /* IE */
    filter: progid:DXImageTransform.Microsoft.gradient(startColorstr='#ADB2B6', endColorstr='#ABAEB3', GradientType=0);
}

Unfortunately this doesn't work correctly in IE as it uses filter for the gradient, which it always paints over the background.

To work around IE's issue you can place the filter and the background image in separate elements. That would obviate the power of CSS3 multiple backgrounds, though, since you can just do layering for all browsers, but that's a trade-off you'll have to make. If you don't need to support versions of IE that don't implement standardized CSS gradients, you have nothing to worry about.

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¹ _{Technically, the background-position and background-repeat declarations apply to both layers here because the gaps are filled in by repeating the values instead of clamped, but since background-position is its initial value and background-repeat doesn't matter for a gradient covering the entire element, it doesn't matter too much. The details of how layered background declarations are handled can be found here.}

CSS - Transparency Gradient on an Image

My solution to my problem is to simply state that this is not possible with the current technology. An alternative option would be to use a simple transparency gradient. Until A better solution arrises this is what I will end up doing.

background: linear-gradient(to bottom, rgba(0,0,0,1) 0%,rgba(0,0,0,0) 100%);

Cut a hole in a background gradient

I was able to figure it out by adding mask-image.

body { background: purple; }
.container { width: 200px; }

.spinner {
  height: 0px;
  width: 100%;
  padding-bottom: calc(100% - 20px);
  margin: 0;
  background: conic-gradient(from 0deg at center, transparent 45deg, orange 360deg);
  clip-path: circle(46% at center);
  -webkit-mask-image: radial-gradient(circle at center, transparent 54%, black 54.8%);
  mask-image: radial-gradient(circle at center, transparent 54%, black 54.8%);
}

<div class="container">
  <div class="spinner"></div>
</div>

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