Hi, I've created an application with the option to start on Windows startup.
First I did this via the Registry, like this:
private void RunOnStartup(bool RunOnStartup) {
Microsoft.Win32.RegistryKey key = Microsoft.Win32.Registry.CurrentUser.OpenSubKey("SOFTWARE\\Microsoft\\Windows\\CurrentVersion\\Run", true);
if (RunOnStartup) {
key.SetValue(ProgramTitle, System.Windows.Forms.Application.ExecutablePath.ToString());
} else {
key.DeleteValue(ProgramTitle, false);
}
}
And this worked, but not correctly. It started the .exe but with the same behavior as it was a new one with the default 'config.xml' it needs. Which is obviously wrong.
I did not manage to find out what was wrong, so I tried it differently: create a shortcut in the Startup folder. Couldn't go wrong I figured, I mean, it's just a shortcut right?
I used this code:
private void RunOnStartup(bool RunOnStartup) {
string startup = Environment.GetFolderPath(Environment.SpecialFolder.Startup) + "\\"+ProgramTitle+".url";
if (RunOnStartup) {
using (StreamWriter writer = new StreamWriter(startup)) {
string app = System.Reflection.Assembly.GetExecutingAssembly().Location;
writer.WriteLine("[InternetShortcut]");
writer.WriteLine("URL=file:///" + app);
writer.WriteLine("IconIndex=0");
string icon = app.Replace('\\', '/');
writer.WriteLine("IconFile=" + icon);
writer.Flush();
}
} else {
if (File.Exists(startup)) {
File.Delete(startup);
}
}
}
And this worked as well, it started, but with the same behavior.
So my question is, does anyone have any idea how this happens? Any help is much appreciated!
Thanks
