WPF中的C#4.0 GetWindowRect [英] C#4.0 GetWindowRect in wpf
本文介绍了WPF中的C#4.0 GetWindowRect的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想获取wpf接口的位置.此代码可以在c#2.0中工作,但是在c#4.0中报告错误.这是代码.
I want to get the position of my wpf interface.This code can work in c#2.0,but Report an error in c#4.0.Here is the code.
[DllImport("user32.dll")]
[return: MarshalAs(UnmanagedType.Bool)]
static extern bool GetWindowRect(HandleRef hWnd, out RECT lpRect);
[StructLayout(LayoutKind.Sequential)]
public struct RECT
{
public int Left;
public int Top;
public int Right;
public int Bottom;
}
Rectangle myRect = new Rectangle();
private void button1_Click(object sender, System.EventArgs e)
{
RECT rct;
if(!GetWindowRect(new HandleRef(this, this.Handle), out rct )) //Here is the error
{
MessageBox.Show("ERROR");
return;
}
MessageBox.Show( rct.ToString() );
myRect.X = rct.Left;
myRect.Y = rct.Top;
myRect.Width = rct.Right - rct.Left + 1;
myRect.Height = rct.Bottom - rct.Top + 1;
}
推荐答案
您应该使用
WindowInteropHelper();
像这样(用于wpf)获取句柄
to get the handle, like this (for wpf)
if(!GetWindowRect(new HandleRef(this, new WindowInteropHelper(this).Handle), out rct ))
you can find more documentation on : https://msdn.microsoft.com/fr-fr/library/system.windows.interop.windowinterophelper%28v=vs.110%29.aspx?f=255&MSPPError=-2147217396
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