Question

I’m making a program that is using Windows Shell Integration, and the registry changes I do are these:
For example, for .txt, I see that the value for HKEY_CLASSES_ROOT\.txt > (Default) is txtfile. I add to HKEY_CLASSES_ROOT\txtfile\shell the key myprogram > (Default) with the value Open with MYPROGRAM. to myprogram I add command > (Default) with the value *path-to-my-program* %1. Now when I right click a .txt file there is an option to open it with my program.

But when I do that with multiple .txt files Windows opens my program many times with each time another file as argument. But I want to open my program one time with all the files as many arguments. Is there an option to do that with changing stuff in registry?

If not, I also could not find a way to make a program that can be opened multiple times and combine all of them to one, so I can also do it that way if someone can help me with it. I’m making this program with C#, by the way.

Answer 1

You have to make your application “Single Instance”.
Something like this should do the trick:
(untested code, just for reference)

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

namespace YourApp
{
    class Program
    {
        [DllImport("user32.dll")]
        [return: MarshalAs(UnmanagedType.Bool)]
        static extern bool SetForegroundWindow(IntPtr hWnd);

        [STAThread]
        static void Main()
        {
            bool createdNew = true;
            using (Mutex mutex = new Mutex(true, "MyApplicationName", out createdNew))
            {
                if (createdNew)
                {
                    Application.EnableVisualStyles();
                    Application.SetCompatibleTextRenderingDefault(false);
                    Form1 frm = new Form1();
                    frm.SetNewData("send command line here");
                    Application.Run(frm);
                }
                else
                {
                    Process current = Process.GetCurrentProcess();
                    foreach (Process process in Process.GetProcessesByName(current.ProcessName))
                    {
                        if (process.Id != current.Id)
                        {
                            SetForegroundWindow(process.MainWindowHandle);
                            // send message to that form or use .Net remoting
                            break;
                        }
                    }
                }
            }
        }
    }
}

For a better example see this CodeProject solution.