在NSImageView中获取NSImage的界限

Question

我有一个占用窗口全部范围的NSImageView.图像视图没有边框,其设置显示在左下角.因此,这意味着视图的原点与实际图像的原点匹配,无论窗口的大小如何调整.

此外,图像比我在屏幕上完全适合的尺寸大得多.所以我也将imageview设置为按比例缩小图像的大小.但是,我似乎无法在任何地方找到这种比例因子.

我的最终目标是将鼠标按下事件映射到实际图像坐标.要做到这一点,我想我还需要一条信息......显示的NSImage实际上有多大.

如果我看一下[imageView bounds],我会得到图像视图的边界矩形,它通常会比图像大.

Answer 1

我认为这给了你你所需要的：

NSRect imageRect = [imageView.cell drawingRectForBounds: imageView.bounds];

它返回视图中图像原点的偏移量及其大小。

对于您重新映射鼠标坐标的最终目标，您的自定义视图类上的类似内容应该可以工作......

- (void)mouseUp:(NSEvent *)event
{
    NSPoint eventLocation = [event locationInWindow];    
    NSPoint location = [self convertPoint: eventLocation fromView: nil];

    NSRect drawingRect = [self.cell drawingRectForBounds:self.bounds];

    location.x -= drawingRect.origin.x;
    location.y -= drawingRect.origin.y;

    NSSize frameSize = drawingRect.size;
    float frameAspect = frameSize.width/frameSize.height;

    NSSize imageSize = self.image.size;
    float imageAspect = imageSize.width/imageSize.height;

    float scaleFactor = 1.0f;

    if(imageAspect > frameAspect) {

        ///in this case image.width == frame.width
        scaleFactor = imageSize.width / frameSize.width;

        float imageHeightinFrame = imageSize.height / scaleFactor;

        float imageOffsetInFrame = (frameSize.height - imageHeightinFrame)/2;

        location.y -= imageOffsetInFrame;

    } else {
        ///in this case image.height == frame.height
        scaleFactor = imageSize.height / frameSize.height;

        float imageWidthinFrame = imageSize.width / scaleFactor;

        float imageOffsetInFrame = (frameSize.width - imageWidthinFrame)/2;

        location.x -= imageOffsetInFrame;
    }

    location.x *= scaleFactor;
    location.y *= scaleFactor;

    //do something with you newly calculated mouse location    
}