Opening process and changing window position

Question

I want to open from c# an application (standalone flashplayer) and set it position to (0,0) on the screen. How can I do this? So far I\'ve managed to open flashplayer:

Answer 1

Once you start the Process, its MainWindowHandle property should be set to some Windows handle that can be used for manipulating with the main window of the started application. I don't think there is a way to move it directly using .NET API, but you can use the MoveWindow API function via P/Invoke.

Here are some links where you can find more information:

• MainWindowHandle property of Process at MSDN
• MoveWindow API function at pinvoke.net

Answer 2

thanks guys, it's working now! :)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;
using System.Runtime.InteropServices;
using System.Threading;

namespace swflauncher
{
    class Program
    {
        static void Main(string[] args)
        {
            Process flash = new Process();
            flash.StartInfo.WindowStyle = ProcessWindowStyle.Normal;

            flash.StartInfo.FileName = "D:\\development\\flex4\\runtimes\\player\\10\\win\\FlashPlayer.exe";
            flash.Start();
            Thread.Sleep(100);

            IntPtr id = flash.MainWindowHandle;
            Console.Write(id);
            Program.MoveWindow(flash.MainWindowHandle, 0, 0, 500, 500, true);
        }

        [DllImport("user32.dll", SetLastError = true)]
        internal static extern bool MoveWindow(IntPtr hWnd, int X, int Y, int nWidth, int nHeight, bool bRepaint);


    }
}

Answer 3

Try SetWindowPos as described here. This page shows how to call it from C#.