问题描述
我正在使用 HTML div 元素 javascript 中基于角的调整大小.
I am working with HTML div element corner based resize in javascript.
基于角的调整大小在选择框 div 未旋转(即 rotate(0 deg))时完美运行.
Corner based resize working perfect when selection box div is unrotated (i.e. rotate(0 deg)).
旋转后，当左上角需要保持在同一位置时，我使用右下角作为手柄来调整选择框div的大小.p>
就像有四个手柄一样，调整大小时对角需要保持在同一位置.
After rotate, i am using bottom-right corner as handle to resize the selection box div, when top-left corner needs to remain in same position.
但问题是，选择框左上角角不保持在同一位置.
As like having four handles, with opposite corner needs to remain in same position when resize.
我需要为旋转后基于角的调整大小计算平移 x 和 y 值.
But the problem is, the selection box top-left corner not remains in same position.
scale, center_x, center_y, previous_x, previous_y, width, height, previous_width, previous_height, diff_width, diff_height, radian_angle, rotated_degree.
I need to calculate translate x and y value for corner based resize after rotate.
scale,center_x,center_y,previous_x,previous_y,width,height,previous_width,previous_height,diff_width,diff_height,radian_angle,rotated_degree.
用于旋转后调整大小的示例代码:
switch (handle) {
case 'bottom-right':
x = previous_x - ? ; //logic behind after rotated some angles..?
y = previous_y + ? ;
break;
case 'bottom-left':
x = ? ;
y = ? ;
break;
case 'top-left':
x = ? ;
y = ? ;
break;
case 'top-right':
x = ? ;
y = ? ;
break;
default:
}
)调整大小时，基于角的调整大小代码可以完美运行:
):
case 'bottom-right':
x = previous_x;
y = previous_y;
break;
case 'bottom-left':
x = previous_x - diff_width;
y = previous_y;
break;
如何计算 translate x &调整大小时旋转后的y值以获得固定角.?
How to calculate translate x & y value after rotate to get fixed corner when resize.?
推荐答案
不知道你是否还需要帮助.但我认为你可以用三角函数(余弦、正弦和你的旋转角度)来达到它，因为你的旋转画了一个圆.
I don't know if you still need help. But I think you can reach it with Trigonometry (cosinus, sinus and your rotation angle) because your rotation draw a circle.
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