- 1 year ago
Need to make the window active
To make a window active programmatically in C#, you can use the SetActiveWindow
function from the User32
library. Here's an example of how you can achieve this:
csharpusing System;
using System.Runtime.InteropServices;
public class WindowActivator
{
[DllImport("user32.dll")]
public static extern IntPtr SetActiveWindow(IntPtr hWnd);
[DllImport("user32.dll")]
public static extern IntPtr FindWindow(string lpClassName, string lpWindowName);
public static void ActivateWindow(string windowTitle)
{
IntPtr windowHandle = FindWindow(null, windowTitle);
if (windowHandle != IntPtr.Zero)
{
SetActiveWindow(windowHandle);
}
}
}
public class Program
{
public static void Main()
{
string windowTitle = "Window Title";
WindowActivator.ActivateWindow(windowTitle);
}
}
In the example above, the WindowActivator
class contains the SetActiveWindow
function imported from the User32
library. The FindWindow
function is used to retrieve the handle of the window based on its title. The ActivateWindow
method takes the window title as an argument, finds the window handle, and then calls SetActiveWindow
to make the window active.
Replace "Window Title"
with the title of the window you want to activate. If the window is found and successfully activated, it will become the active window.
Please note that you need to import the user32.dll
library and the necessary methods using the DllImport
attribute. The code provided is for Windows platforms, and it assumes you are using C# for your application development.