如何从给定的xml数据计算MS Office Powerpoint形状的旋转值 [英] how to calculate the rotation value for the MS Office Powerpoint Shapes from the xml data given

问题描述

我想从给定的OOXML数据中获得适当的旋转值以绘制MS Office Powerpoint 2007文件的形状，如下所示:

I want to get the proper rotation value to draw the shape of the MS office Powerpoint 2007 file from the given OOXML data as below :

<p:sp>
            <p:nvSpPr>
                <p:cNvPr id="3" name="Rectangle 66" />
                <p:cNvSpPr>
                    <a:spLocks noChangeArrowheads="1" />
                </p:cNvSpPr>
                <p:nvPr />
            </p:nvSpPr>
            <p:spPr bwMode="auto">
                ***<a:xfrm rot="5400000">***
                    <a:off x="2443049" y="-1042472" />
                    <a:ext cx="304800" cy="4419600" />
                </a:xfrm>
                <a:prstGeom prst="rect">
                    <a:avLst />
                </a:prstGeom>
                <a:ln>
                    <a:headEnd />
                    <a:tailEnd />
                </a:ln>
            </p:spPr>

旋转值指定为"xfrm rot = 5400000".考虑到这一点，必须计算其适当的值，因此必须处理形状的高度和宽度以绘制形状.

The value of rotation is given as "xfrm rot = 5400000". Considering this, its proper value has to be calculated and accordingly the height and width of the shape has to handled to draw the shape.

推荐答案

将其除以60000得到旋转角度.在这种情况下，5400000/60000 = 90度.它是就地旋转，这意味着它在中心X和中心Y上旋转.

Divide it by 60000 to get the rotation angle. In this case, 5400000/60000=90 degrees. It is an in-place rotation, meaning it rotates on Center X and Center Y.

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