将坐标从iPhone屏幕尺寸转换为iPad屏幕尺寸 [英] Convert coordinates from iPhone screen size to iPad screen size

问题描述

我有一个自定义UIView，其中包含在视图的drawRect函数中绘制的交互式绘图。在iPad上，绘图尺寸为1024 x 768.对于iPhone，我使用CGContextScaleCTM缩小iPhone的绘图。为了在我的绘图上保持适当的宽高比，我将视图缩小到480 x 360并将视图的y值设置为-20（从视图的顶部和底部有效裁剪20个像素，这很好）。现在一切看起来都很正确，但是我需要将触摸坐标从iPhone转换为iPad坐标，以便我的视图的交互部分能够正常工作。如果我将uiview 320设为高并使用

I have a custom UIView that contains an interactive drawing that is drawn in the drawRect function of the view. On the iPad the drawing size is 1024 x 768. For the iPhone I shrink the drawing for the iPhone using CGContextScaleCTM. To keep the proper aspect ratio on my drawings I shrink the view to 480 x 360 and set the y value of the view to -20 (effectively cropping 20 pixels off the top and bottom of the View, which is fine). Everything looks correct now, but I need to convert the touch coordinates from iPhone to iPad coordinates for the interactive portions of my view to work. If I make the uiview 320 high and use

point.y *=768/320 

用于转换y值的位置是正确的（但我的绘图是扭曲的）我已经做了一些测试硬编码点所以我知道这个应该工作，但我很难让数学与作物一起工作。以下是我到目前为止：

for converting the y value the locations are correct (but my drawing is distorted) I've done some tests hard coding point in so I know this should work but I'm having a hard time getting the math to work with the crop. Here is what I have so far:

CGPoint point = [[touches anyObject] locationInView:self.view];
[self endTouch:&point];
NSLog(@"true point: %@",NSStringFromCGPoint(point));
if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone) {
point.x *= 1024/480;
point.y *= 768/360;
NSLog(@"corrected point: %@",NSStringFromCGPoint(point));
}

有人可以帮我做这个转换吗？谢谢！

Can anybody help me with doing this conversion? Thanks!

推荐答案

1024/480 和 768/360 执行整数除法。两个操作的结果是 2 ，但你真的想要 2.133333 。用以下内容替换代码中的相关行：

1024/480 and 768/360 perform integer division. The result of both operations is 2, but you really want 2.133333. Replace the relevant lines in your code with the following:

point.x *= 1024.0/480.0;
point.y *= 768.0/360.0;

这将返回您正在寻找的值并缩放您的点 x 和 y 相应的值。

This will return the value you're looking for and scale your point's x and y values accordingly.

您可能会更好地服务（就可读性而言） ）用 #define 宏替换这些文字。在代码的顶部，输入以下内容：

You might be better served (in terms of readability) by replacing these literals with a #define macro. At the top of your code, put the following:

#define SCALING_FACTOR_X (1024.0/480.0)
#define SCALING_FACTOR_Y (768.0/360.0)

然后修改你的代码使用：

and then modify your code to use that:

point.x *= SCALING_FACTOR_X;
point.y *= SCALING_FACTOR_Y;

这样，你会更清楚地知道你在做什么。

That way, it will be more clear as to what you're actually doing.

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