Adding my program to contextmenu

This is a discussion on Adding my program to contextmenu within the C# Programming forums, part of the General Programming Boards category; To add my program into the context menu (when the user right click on excel file). I used the following ...

  1. #1
    Registered User
    Join Date
    Feb 2012
    Posts
    1

    Adding my program to contextmenu

    To add my program into the context menu (when the user right click on excel file). I used the following code:

    Code:
    public static bool AddContextMenuItem(string Extension, string MenuName, string MenuDescription, string MenuCommand)
            {
                bool ret = false;
                RegistryKey rkey = Registry.ClassesRoot.OpenSubKey(Extension);
    
                if (rkey != null)
                {
                    string extstring = rkey.GetValue("").ToString();
    
                    rkey.Close();
    
                    if (extstring != null)
                    {
                        if (extstring.Length > 0)
                        {
                            rkey = Registry.ClassesRoot.OpenSubKey(extstring, true);
    
                            if (rkey != null)
                            {
                                string strkey = "shell\\" + MenuName + "\\command";
    
                                RegistryKey subky = rkey.CreateSubKey(strkey);
    
                                if (subky != null)
                                {
                                    subky.SetValue("", MenuCommand);
                                    subky.Close();
                                    subky = rkey.OpenSubKey("shell\\" + MenuName, true);
                                    if (subky != null)
                                    {
                                        subky.SetValue("", MenuDescription);
                                        subky.Close();
                                    }
                                    ret = true;
                                }
                                rkey.Close();
                            }
                        }
                    }
                }
    
                return ret;
            }
    and:

    Code:
    public static bool ProcessCommand(string[] args)
            {
                
                if (args.Length == 0 || string.Compare(args[0], "-register", true) == 0)
                {
                    
                    string menuCommand = string.Format("\"{0}\" \"%L\"", Application.ExecutablePath);                
    
                    //frmMain.Register(Program.FileType, Program.KeyName, Program.MenuText, menuCommand);                
                    
                    AddContextMenuItem(FileType, KeyName, MenuText, menuCommand);
                    
                    MessageBox.Show(string.Format("The {0} shell extension was registered.", Program.KeyName), Program.KeyName);
    
                    return true;
                }          
                
                return false;
            }
    I used the following function to retrieve the excel file path has been rightclick choose to load the program:

    Code:
    public static void OpenExcelFileWithEasyForm(string filePath)
            {
                try
                {                
                    string excelFilePath = Path.Combine(Path.GetDirectoryName(filePath), string.Format("{0} (EasyForm){1}", Path.GetFileNameWithoutExtension(filePath), Path.GetExtension(filePath)));
                                 
                    
                }
                catch (Exception ex)
                {
                    MessageBox.Show(string.Format("An error occurred: {0}", ex.Message), Program.KeyName);
                    return;
                }
            }
    In main():

    Code:
    if (!ProcessCommand(args))
                {
                    OpenExcelFileWithEasyForm(args[0]);
                }
    But I still do not load the excel file into my program by selecting the program from ContextMenu (right click on excel file and choose program).
    Although if select the path to excel file from the program, the data from excel file is loaded into the program.

    I'm still missing something at function OpenExcelFileWithEasyForm(). Has anyone ever done this, please guide me to complete this function to load the excel file into my program when selecting programs in ContextMenu.

    Thanks

  2. #2
    Gawking at stupidity
    Join Date
    Jul 2004
    Location
    Oregon, USA
    Posts
    3,164
    So in OpenExcelFileWithEasyForm() you've created the path to your EasyForm, but you're not doing anything with it. If you want to open the file using the default program that's registered with Windows for that file extension, you can just do this:
    Code:
    System.Diagnostics.Process.Start(excelFilePath);
    If you understand what you're doing, you're not learning anything.

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