gradient

I'm trying to make a similar status bar and navigation bar gradient as Google Now. Image Reference: Rectangular area indicated as below After trying the below option on Android Marshmallow, <item name="android:windowTranslucentNavigation">true</item> <item name="android:windowTranslucentStatus">true</item> I get the below behaviour Image Reference: Can anyone suggest me how to get gradient on both these ? Is that a gradient or is that a shadow ? Bryan It is a gradient being used as a scrim . To achieve this effect, the first thing you have to do is remove the semi-transparent underlay for the

My goal is to reproduce this plot, and my problem is reproducing the gradient fill in each bar. ADDED after COMMENT The good comment of @PavoDive directs us to a question that basically says "You can't do it with ggplot2 and furthermore it is a bad idea even if you could do it." Agreed as to it being a poor graphical choice, but for didactic purposes I wanted to re-create the original and then show improvements. So, is there a programming solution nevertheless? With the data that follows the ggplot code I have gotten close, other than the consistent gradient coloring that is the same for each

Closed. This question is off-topic. It is not currently accepting answers. Learn more . Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . Simply put, I need to put a gradient on a dynamically sized circle with a transparent background. Is there a way to recreate the following image in CSS/HTML ? As far as I can tell, neither radial gradients nor border-images are able to construct this shape. The key is that the center of the circle needs to be transparent, as I can not "fake" a torus by filling the center with white. Update: This

I'm a beginner at SVG, but I'd like to learn some techniques. To be short, is there a simple way to create something like this? I was thinking about creating a polar gradient and then clipping it: But how do I generate a polar gradient? Even if there's no native method, maybe it could be made with a simple linear gradient and then using some rectangular-polar coordinate transformation. Is there a way to do so? So this is the solution I developed: <?xml version="1.0" standalone="no"?> <!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 1.1//EN" "http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd"> <svg viewBox

I have not really done much programming with Core Graphics. And I tend to stick with QuartzCore now that it does a lot of what I need through the layer property :) However, I have a UIView, which is currently gradient. I'd like to add rounded corners to this UIView and the layer property does not do this when I draw the gradient: - (void)drawRect:(CGRect)rect { CGContextRef currentContext = UIGraphicsGetCurrentContext(); CGGradientRef glossGradient; CGColorSpaceRef rgbColorspace; size_t num_locations = 2; CGFloat locations[2] = { 0.0, 1.0 }; CGFloat components[8] = { 1.0, 1.0, 1.0, 0.95, //

I’m making a set of buttons which use dynamic gradients. I’ve taken care of Firefox 3.6+ and WebKit by using their proprietary CSS extensions and all I need to do is support Opera, iOS and IE9 by using background-image: url("gradient.svg"). This is relatively easy, I made an SVG file, linked it and got it working. However, I’m making a set so I need at least 6 gradients. When I normally do it in images, I create a sprite for fast HTTP access. I’m not sure how to achieve this in SVG – can I use one file and access different parts of its XML by using #identifiers, like XBL does? My current SVG:

I know there is -webkit-gradient and -moz-linear-gradient . But what is the standard way of defining a gradient? Like there is: -webkit-border-radius and -moz-border-radius and the standard is meant to be border-radius . An update for 2011, the Mozilla syntax is now the 'official' one, adopted by the CSS3 Image Values and Replaced Content Working Draft . Webkit has been updated to use this syntax too, and this has now been incorporated into the latest versions of Chrome and Safari. Kyle Not yet, the two examples you provided are the only coded gradients available as yet, as far as I know. Most