Develop Reference

Registry Startup not working windows application

I created windows application ,I want to start my application on windows startup
for that i written following code in installer class. but when i am checking registry using regedit i didnt get registry value. and my application not working.
public override void Commit(IDictionary savedState)
{
base.Commit(savedState);
try
{
RegistryKey add = Registry.CurrentUser.OpenSubKey("SOFTWARE\\Microsoft\\Windows\\CurrentVersion\\Run", true);
add.SetValue("ToposcreenServer", "\"" + Application.ExecutablePath.ToString() + "\"");
RegistryKey key = Registry.LocalMachine.CreateSubKey(#"SOFTWARE\Wow6432Node\Microsoft\Windows\CurrentVersion\Uninstall\{70E25B31-99A9-474C-8990-CE28FBCEAAD1}", RegistryKeyPermissionCheck.Default);
if (key != null)
{
key.SetValue("SystemComponent", 1, RegistryValueKind.DWord);
key.Close();
}
Directory.SetCurrentDirectory(Path.GetDirectoryName(Assembly.GetExecutingAssembly().Location));
Process.Start(Path.GetDirectoryName(Assembly.GetExecutingAssembly().Location) + "\\ToposcreenServer.exe");
GLobalclass.WriteLog("Installer Executed");
}
catch (Exception ex)
{
GLobalclass.WriteLog("Installer Error :" + ex.Message);
}
}

If this is an Everyone install then that code won't write to HKCU of the installing user because the code is running with the System account credentials, not the installing user's.
Anyway, you don't need code to set the Run key. Go to the registry view in the IDE and add registry folders to get to that Run key in HKCU. Then add an item with the Nama ToposcreenSaver and the value [TARGETDIR]my.exe assuming your executable is in the Application Folder in the File System view. It's possible that it won't run anyway if it requires elevation on a UAC system.
(If this code is really in an installer class, it's also not clear why you're using Application and ExecuteablePath because an installer class is a Dll being called from an msiexec.exe process, and is nothing at all to do with whatever executable you want to run. Surely it's the name of an executable you are installing?)
You don't need to set SystemComponent in the registry key. That registry key may not be there at the time your custom action runs, and what you should really do is open your MSI file with Orca and add ARPSYSTEMCOMPONENT to the Property table, give it a value of 1.
https://msdn.microsoft.com/en-us/library/windows/desktop/aa367750(v=vs.85).aspx
If the app really is a conventional screensave this might be the best way to do it:
http://www.advancedinstaller.com/user-guide/qa-install-screensaver.html

You need to check that add is not null, as per https://msdn.microsoft.com/en-us/library/xthy8s8d(v=vs.110).aspx .

How can I set this registry value for my User from my installer

The problem from
https://stackoverflow.com/a/37859812/4878558
I need to set Registry value for current user, who launch the install up. Since install going for system mode - I don't know anything about current user
Also my code giving 'System.UnauthorizedAccessException'
SecurityIdentifier sID = WindowsIdentity.GetCurrent().User;
var subKey = Registry.Users.OpenSubKey(sID + "\\SOFTWARE\\Microsoft\\Windows\\CurrentVersion\\Run");
subKey.SetValue("test", "test");
enter code here

As Ripple and I have both commented, there's no need for code. Go to the Registry view in the setup project, right-click on Software under HKEY_CURRENT_USER and add the key Microsoft, then Windows, the CurrentVersion, then Run, adding each key.
Then in the Run key view, right-click in the Name, View pane on the right and add new string value, the name being your name. The value, I assume, is the path to your exe, and (assuming it's in the Application folder) make the value [TARGETDIR]my.exe.
If your install is an "Everyone" install then there is a perfectly good reason why it cannot work. This is nothing to do with the code. In an Everyone install that custom action code is running with the System account (NOT the installing user) so you are trying to create a run key for the system account.

Here is how to write autostartup options:
const string AutorunRegistryKey = #"HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Run";
Registry.SetValue(AutorunRegistryKey, <AppName>, <PathToApplication>);
If you want to remove it from autostartup:
const string AutorunRelativePath = #"Software\Microsoft\Windows\CurrentVersion\Run\";
var key = Registry.CurrentUser.OpenSubKey(AutorunRelativePath, true);
if (key != null)
{
key.DeleteValue(<AppName>, false);
key.Close();
}

How can I get my app to run from the Start menu?

If you type "regedit" in the Start menu's edit box and mash the Enter key, Registry Editor will be invoked. The same is true for "cmd" and the Command Line, and doubtless several other apps.
How can I get my app to respond the same way, so that if the user enters "Platypus" in the Start menu edit box, Platypus.exe will be invoked?
Does it require manipulation of the Registry / adding an entry somewhere there, and if so, just what key and value needs to be added?
I would be satisfied with the user needing to run the app manually once (2-clicking its icon; it's a Winforms app), at which time startup code (no pun intended) would do whatever was necessary to make the app henceforth Startsmartable (Windows key, "Platypus", to start the app).
I know that it's just as easy/easier for the user to simply 2-click a desktop icon when they want to run the app, but this particular functionality is not my idea, so complaints about the oddity of this question would be to no avail.
UPDATE
I added the code recommended by Chandan (with my executable's name):
public static void AddToStartup()
{
using (RegistryKey startup = Registry.CurrentUser.OpenSubKey("SOFTWARE\\Microsoft\\Windows\\CurrentVersion\\Run", true))
{
startup.SetValue("RoboReporter", "\"" + System.Windows.Forms.Application.ExecutablePath + "\"");
}
}
...called it from the main form's load event:
private void FormRoboReporter_Load(object sender, EventArgs e)
{
RoboReporterConstsAndUtils.AddToStartup();
}
...shut down the app, went to the Start menu and entered the program's name ("RoboReporter"), and all it did was bring up search results of related file names.
UPDATE 2
What it does do is cause my app to run whenever the computer is restarted. That's not what I want. The code above adds an entry to HKEY_CURRENT_USER.Software.Microsoft.Windows.CurrentVersion.Run as can be seen here (along with a couple of other entries that predated it):
I don't want the app to start up every time the computer restarts, so I removed the entry. The question remains: how can I make the app runnable from the Start menu?

You can add your application's parent directory's path to the environment variable called PATH.
string pathvar = Environment.GetEnvironmentVariable("PATH", EnvironmentVariableTarget.Machine);
Environment.SetEnvironmentVariable("PATH", pathvar + ";" + Application.StartupPath + "\\", EnvironmentVariableTarget.Machine);
(Note that the paths added to this variable should end with a backslash \, and each path is separated by a semicolon ;)
Adding the parent directory's path to the environment variable will make all it's contents quickly accessible from the Start Menu's search field, from Run and from CMD.
You can also change EnvironmentVariableTarget.Machine to EnvironmentVariableTarget.User to modify the variable for the current user only.
EDIT:
A note: Setting a variable for the entire machine (by using EnvironmentVariableTarget.Machine) seems to require elevated privileges when done from one's application.

you might want to run this
public static void AddToStartup()
{
using (RegistryKey startup = Registry.CurrentUser.OpenSubKey("SOFTWARE\\Microsoft\\Windows\\CurrentVersion\\Run", true))
{
startup.SetValue("Name_of_your_Program", "\"" + Application.ExecutablePath + "\"");
}
}

application won't start on startup after adding a notifyicon

my app was working ok and it would execute on startup before.
I added a notify icon and in my code,there are some places that this icon changes.I added all required icons in the root folder of my app,and everything is working fine with the icons,except the startup boot of my app.
I can see my app's address in the "run" part of the registry(I mean everything is the same as when my app booted at startup properly).but my app won't run at startup anymore.
any advice on my matter?
PS:I thought I should explain my work a little bit and I wrote a little piece of app that has the exact same problem
public Icon[] icons = new Icon[2] { new Icon("icon1.ico"), new Icon("icon2.ico") };
public int counter = 0;
private void button1_Click(object sender, EventArgs e)
{
notifyIcon1.Visible = true;
timer1.Start();
}
private void timer1_Tick(object sender, EventArgs e)
{
counter %= 2;
notifyIcon1.Icon = icons[counter];
counter++;
As you can see,the app changes the icon of the notifyicon in every tick.with this code,the app won't run at startup.but if I remove the iconchanging feature of the app,it will actually run at startup

This requires psychic debugging, I'll guess that you are loading these icons using their relative path name. Something like new Icon("foo.ico").
This can only work correctly if the default working directory of your program is set where you hope it will be. It usually is, certainly when you start your program from Visual Studio or start it from a desktop shortcut. But not when you added it to the Run registry key. Environment.CurrentDirectory will be set elsewhere, typically the Windows directory.
You must always use the full path name of files. An easy way to get that path is:
var home = System.IO.Path.GetDirectoryName(System.Reflection.Assembly.GetEntryAssembly().Location);
var path = System.IO.Path.Combine(home, "foo.ico");
var icon = new Icon(path);
But there's certainly a better way than storing icons as files, you can embed them in your program. Project + Properties, Resources tab. Click the arrow on the Add Resource button, Add Existing File and navigate to your .ico file. Now the icon is embedded in your program, you'll never lose track of it and can't forget to copy it when you deploy your program on another machine. And the code is simpler as well:
var icon = Properties.Resources.foo;

How to get the (.lnk) shortcut filepath in a program which started by the shortcut?

I have a c# program which open *.postfix file.
If a user runs a (.lnk)shortcut which points to my type of file, my program will open the target.
So, how could my program know it is started by a (.lnk)shortcut (and get it's file path)?
In some circumstances,i need to replace the .lnk file.
Thanks!
Edited
First, thanks to guys who answered my question.
By following #Anders answer, i find out my problem lays here.
I made some changes to windows registry, so browser knows to throw customized protocol string to certain program.
some thing like this..
[InternetShortcut]
URL=myProtocol://abcdefg.....
That's maybe why i lost lpTitle. :(
I'm going to try this way:
Whenever my program invoked, of course fed with %1, program checks current opened explorer(Window), and try to get it's current path with IWebBrowserApp. With that path and desktop of course, scan and analyze *.lnk to determine which one to replace.
I think this will probably work, but not be sure. I will try.
continued

In native code you can call GetStartupInfo, if the STARTF_TITLEISLINKNAME bit is set in STARTUPINFO.dwFlags then the path to the .lnk is in STARTUPINFO.lpTitle. I don't know if there is a .NET way to get this info, you probably have to P/Invoke...

You don't. There's no way to do it. End of story.
So this has been brought to my attention due to a recent downvote. There's an accepted answer showing an idea that gets the path to the launching shortcut most of the time. However my answer is to the whole. OP wants the link to the shortcut so he can change it. That is what can't be done most of the time.
Most likely case is the shortcut file exists in the start menu but is unwritable. However other cases involve the shortcut coming from another launching application that didn't even read it from a disk but from a database (I've seen a lot of corporate level restricted application launch tools). I also have a program that launches programs from shortcuts not via IShellLink but by parsing the .lnk file (because it must not start COM for reasons) and launching the program contained. It doesn't pass STARTF_TITLEISLINKNAME because it's passing an actual title.

If you're using Visual Studio Setup Project to build an installer and do the file type association, you should follow these instructions http://www.dreamincode.net/forums/topic/58005-file-associations-in-visual-studio/
Open up your solution in Visual studio.
Add a Setup Project to your solution by file , add project,New project, Setup & Deployment projects,Setup project
Right-click on your setup project in the "Solution Explorer" window,Select view,then select file types.
you'll see the "file types" window displayed in Visual studio.At the top of the window will be "File types on target machine"
Right-click on "File types on target machine".the menu will pop up with Add "file type" Click on this.
you'll see "New document Type#1" added,and "&open"underneath it.
The "new document type#1" can be anything you want - change it to something descriptive.although the user never sees this,never use something common- be as unique as possible,Because you can overlay current file associations without even realizing it.For example,you might think"pngfile" might be a useful name- but using that will now send all"*.png" files to your application,instead of to an image viewer.A good practice maybe "YourCompantName.Filetype",where your company name is your name of your company's name, and "Filetype" is a descriptive text of your file.
In the "properties" window for your new type,you will need to change a few properties.:
Command:Change to the application that you want to run.If you click on the "..." and you will proberly want to locate and use the "primary Output..." File
Description: This is the description of the file type(if it doesn't describe it's self"
Extensions:This your list of extensions for you chosen Program.Separate each one with a ","
Icon:This will associate the icon with your file type,This shows up in the window explorer.
Now we move to that "&open ".This is an action that is available if your right-click on the file.The default action("&Open" is currently set as the default) is what happens when you double click on the file.Right click on your "New document type#1" to add actions,but for the moment,lets define our "&open" action
Click on "&Open".You will see in the properties window "Name","Arguments","Verbs". Verb is hidden from the user,but is the key that is stored in the registry.Leave it same as the name,But without the "&".The default for"Arguments" is "%1",Which means to pass the full path and filename to your application.You can add other stuff here as well,if you need to pass flags to your application to do special stuff.All this infomaton is getting passed to your application on the command line,so you'll need to be familiar with the "Environment.CommandLine" object.
If you need to set a different action as your default,just right click on the action and "set as default"
Basically, you'll pass the file path as an argument to your program. Then if it's a console application or Windows Forms , you should check the arguments in Program.Main
static void Main(string[] args)
{
//if file association done with Arguments %1 as per forum post above
//you file path should be in args[0]
string filePath = null;
if(args != null && args.Length > 0)
filePath = args[0];
}
For a WPF application you'll need to handle that in the StartUp event for your Application
void App_Startup(object sender, StartupEventArgs e)
{
string filePath = null;
if ((e.Args != null) && (e.Args.Length > 0))
{
filePath = e.Args[0];
}
}