如何获取CSS旋转元素的实际(而不是原始)高度 [英] How to get actual (not original) height of a CSS-rotated element
问题描述
我需要获得几个不同元素的实际高度(为了精确定制工具提示定位),并且其中一些元素(不是全部)都被旋转。
I need to obtain actual height of several different elements (for the sake of precise custom tooltip positioning), and some of these elements (not all) are rotated. $(elem).outerHeight() returns the original height, instead of the actual displayed height.
下面是一个非常简单的例子: http://jsfiddle.net/NPC42/nhJHE/
Here's the fiddle with a very simple example: http://jsfiddle.net/NPC42/nhJHE/
我看到一个可能的解决方案此答案: http://stackoverflow.com/a/8446228/253974 ,但我仍然希望有一个更简单的方法。
I see a possible solution in this answer: http://stackoverflow.com/a/8446228/253974, but am still hoping there is a simpler way.
推荐答案
把我的高中几何和我强大的图形技巧打掉,我把这个图放在一起。如果您有变量 width
, height
和 rotation
javascript,你可以这样表示高度:
Dusting off my high school geometry and my formidable graphics skills, I put this diagram together. If you have variables width
, height
, and rotation
in javascript, you could express the height this way:
var rotated_height = width * Math.sin(rotation) + height * Math.cos(rotation);
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