问题描述
我正在开发一个使用拖放的 html5 界面.当我拖动一个元素时,目标获得了一个 css 类,这使得它由于 -webkit-animation 而双向旋转.
i am working on a html5 interface wich uses drag and drop. While i am dragging an element, the target gets a css-class, which makes it bidirectionally rotate due to a -webkit-animation.
@-webkit-keyframes pulse {
0% { -webkit-transform: rotate(0deg); }
25% { -webkit-transform:rotate(-10deg); }
75% { -webkit-transform: rotate(10deg); }
100% { -webkit-transform: rotate(0deg); }
}
.drag
{
-webkit-animation-name: pulse;
-webkit-animation-duration: 1s;
-webkit-animation-iteration-count: infinite;
-webkit-animation-timing-function: ease-in-out;
}
当我放下目标时,我希望它采用当前的旋转状态.
When I drop the target, I want it to adopt the current state of rotation.
我的第一个想法是使用 jquery 和 .css('-webkit-transform') 方法检查 css 属性.但是这个方法只返回'none'.
My first thought was to check the css property with jquery and the .css('-webkit-transform') method. But this method just returns 'none'.
所以我的问题是:有没有办法获取通过动画旋转的元素的当前度值?
So my question: Is there a way to get the current degree value of an element which is rotated via animation?
到目前为止谢谢亨德里克
Thanks so far Hendrik
推荐答案
我最近不得不写一个函数来做你想要的!随意使用它:
I recently had to write a function that does exactly what you want! Feel free to use it:
// Parameter element should be a DOM Element object.
// Returns the rotation of the element in degrees.
function getRotationDegrees(element) {
// get the computed style object for the element
var style = window.getComputedStyle(element);
// this string will be in the form 'matrix(a, b, c, d, tx, ty)'
var transformString = style['-webkit-transform']
|| style['-moz-transform']
|| style['transform'] ;
if (!transformString || transformString == 'none')
return 0;
var splits = transformString.split(',');
// parse the string to get a and b
var parenLoc = splits[0].indexOf('(');
var a = parseFloat(splits[0].substr(parenLoc+1));
var b = parseFloat(splits[1]);
// doing atan2 on b, a will give you the angle in radians
var rad = Math.atan2(b, a);
var deg = 180 * rad / Math.PI;
// instead of having values from -180 to 180, get 0 to 360
if (deg < 0) deg += 360;
return deg;
}
希望这会有所帮助!
编辑我更新了代码以使用 matrix3d 字符串,但它仍然只给出 2d 旋转度数(即围绕 Z 轴旋转).
EDIT I updated the code to work with matrix3d strings, but it still only gives the 2d rotation degrees (ie. rotation around the Z axis).
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