在两个不同设备上的同一张图像上绘制相同的圆(不同的屏幕密度)
我在两个具有不同屏幕尺寸和密度的不同设备上有一个图像。
使用画布,我想在设备A图像上画一个圆圈,然后将中心坐标发送到设备B,并在相同位置绘制圆圈,即使同一图像具有不同的大小。
在发送 x 坐标之前,我在设备 A 上执行的操作如下:
float density = getResources().getDisplayMetrics().density;
int width = getWidth();
float inchesLength = width/density;
float scaledXCenter = xCenter / inchesLength;
我对 y 坐标执行相同的操作。
在设备 B 上,我获得相同的参数,并将接收到的坐标乘以英寸长度:
float density = getResources().getDisplayMetrics().density;
int width = getWidth();
float inchesLength = width/density;
float restoredXCenter = scaledXCenter * inchesLength;
我正在 AVD 上对此进行测试。 问题是每个AVD的屏幕密度为1,即使我可以清楚地看到在较小的设备(设备A)上,相同的图像被完全显示,但占用的空间更小!
这种方法在真实设备上工作时应该有效吗?
有没有更好的办法呢?
I have an image on two different devices with different screen size and density.
Using a canvas, I want to draw a circle on device A image, then send the center coordinates to device B, and draw the circle in the same position, even if the same image has a different size.
What I do on device A before sending the the x coordinate is the following:
float density = getResources().getDisplayMetrics().density;
int width = getWidth();
float inchesLength = width/density;
float scaledXCenter = xCenter / inchesLength;
I do the same for the y coordinate.
On device B, I get the same parameters and multiply the received coordinate for the inchesLenght:
float density = getResources().getDisplayMetrics().density;
int width = getWidth();
float inchesLength = width/density;
float restoredXCenter = scaledXCenter * inchesLength;
I am testing this on AVDs.
The problem is that each AVD has a screen density of 1, even if I can clearly see that on the smaller device (device A), the same image is fully displayed, but occupies a smaller space!
Should this approach work when working on real devices?
Is there any better way to do it?
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找到画布的宽度和高度,然后将宽度和高度相除得出比率。该比率应与每个值相乘
这是示例代码
find the width and height of the canvas and by dividing width and height find a ratio. This ratio should be multiplied with each value
Here is the sample code