问题

通过鼠标钩子获取鼠标坐标，还是按照下图：1920 x 1080 显示的，所以是不正确的，应该是按照 1920 / 125%显示

解决

internal class ScreenScale
{
    public const int DESKTOPHORZRES = 118;

    [DllImport("user32.dll")]
    public static extern IntPtr GetDC(IntPtr ptr);

    [DllImport("gdi32.dll")]
    public static extern int GetDeviceCaps(
        IntPtr hdc, // handle to DC
        int nIndex  // index of capability
        );

    [DllImport("user32.dll", EntryPoint = "ReleaseDC")]
    public static extern IntPtr ReleaseDC(IntPtr hWnd, IntPtr hDc); [DllImport("user32.dll")]
    public static extern IntPtr GetDesktopWindow();

    public static float GetScreenScale()
    {
        var hdc = GetDC(GetDesktopWindow());
        int nWidth = GetDeviceCaps(hdc, DESKTOPHORZRES);
        ReleaseDC(IntPtr.Zero, hdc);
        float f_Scale = (float)nWidth / (float)Screen.PrimaryScreen.Bounds.Width;
        return f_Scale;
    }

使用

float scale = ScreenScale.GetScreenScale();

注意

上面代码，必须在下面代码之前调用（之后在任意位置调用都可以），否则无法获取缩放比例：

if (Environment.OSVersion.Version.Major >= 6)
{
    SetProcessDPIAware();
}

上面代码的解释详见：
https://www.malaoshi.top/show_1IX6DMSSMArT.html