How to get a rotated linear gradient svg for use as a background image?

问题

I've seen a few questions dancing around this, so I hope this isn't too redundant. Ideally, I'd like an image/svg+xml which scales to 100% of it's container. Colorzilla gets me a great start with a data:image/svg+xml

<?xml version="1.0" ?>
<svg xmlns="http://www.w3.org/2000/svg" width="100%" height="100%" viewBox="0 0 1 1" preserveAspectRatio="none">
  <linearGradient id="grad-ucgg-generated" gradientUnits="userSpaceOnUse" x1="0%" y1="0%" x2="100%" y2="0%">
    <stop offset="0%" stop-color="#ffffff" stop-opacity="0"/>
    <stop offset="100%" stop-color="#ff0000" stop-opacity="1"/>
  </linearGradient>
  <rect x="0" y="0" width="1" height="1" fill="url(#grad-ucgg-generated)" />
</svg>

Note: the width="100%" height="100%" I'd like to take this gradient and rotate it by, say 65deg The HTML5 canvas API provides a great way for me to build this image and then use .toDataURL() PNG to polyfill IE8 and IE7, but I'd like something scalable for IE9.

So the goal is to replicate this:

background: linear-gradient(bottom, rgba(239, 239, 214,0) 0%, rgba(239, 239, 214,.8) 100%),
linear-gradient(left,  rgba(239, 239, 214,0) 60%,rgba(207, 223, 144,1) 100%),
linear-gradient(right, rgba(239, 239, 214,0) 0%,rgba(239, 239, 214,1) 60%),
linear-gradient(top, rgba(239, 239, 214,0) 60%,#cfdf90 100%);
}

with an image/svg+xml that's 100% width and height.

I did try out http://svg-edit.googlecode.com but the interface was less than intuitive for the types of editing I wanted to do. Thanks!

回答1:

To rotate the gradient you can e.g use the 'gradientTransform' attribute, like this:

<?xml version="1.0" ?>
<svg xmlns="http://www.w3.org/2000/svg" width="100%" height="100%" 
 viewBox="0 0 1 1" preserveAspectRatio="none">
  <linearGradient id="grad-ucgg-generated" gradientUnits="userSpaceOnUse" 
   x1="0%" y1="0%" x2="100%" y2="0%" gradientTransform="rotate(65)">
    <stop offset="0%" stop-color="#ffffff" stop-opacity="0"/>
    <stop offset="100%" stop-color="#ff0000" stop-opacity="1"/>
  </linearGradient>
  <rect x="0" y="0" width="1" height="1" fill="url(#grad-ucgg-generated)" />
</svg>

回答2:

Please note that the gradientTransform attribute rotates the gradient according to it's anchor point at 0,0. To rotate it from the 'center' you need to calculate the proper percentages for x1, y1, x2 and y2. A simple PHP example:

// Rotation can be 0 to 360
$pi = $rotation * (pi() / 180);
$coords = array(
    'x1' => round(50 + sin($pi) * 50) . '%',
    'y1' => round(50 + cos($pi) * 50) . '%',
    'x2' => round(50 + sin($pi + pi()) * 50) . '%',
    'y2' => round(50 + cos($pi + pi()) * 50) . '%',
)

回答3:

<linearGradient gradientTransform="rotate(65)">

回答4:

Giel Berkers' solution in Javascript would be:

// angle can be 0 to 360
var anglePI = (angle) * (Math.PI / 180);
var angleCoords = {
    'x1': Math.round(50 + Math.sin(anglePI) * 50) + '%',
    'y1': Math.round(50 + Math.cos(anglePI) * 50) + '%',
    'x2': Math.round(50 + Math.sin(anglePI + Math.PI) * 50) + '%',
    'y2': Math.round(50 + Math.cos(anglePI + Math.PI) * 50) + '%',
}

来源：https://stackoverflow.com/questions/9025678/how-to-get-a-rotated-linear-gradient-svg-for-use-as-a-background-image

标签

css

internet-explorer

svg

canvas

data-uri