Develop Reference

How to Make Multiple Program Files within Same Project in Visual Studio

I am making a simple program that:
Automatically Runs on Startup (Invisibly)
Finishes task within 5-10 seconds (Invisibly)
Exits
I am a beginner and have one challenge here. There are two programs. One which user can open to change settings (Windows Forms) and another that runs on startup, finishes task & exit. How can I make two programs in a single project?
Also, where to save configuration so that both the programs can read/write it?
Thanks for help in advance.
Update: The program is basically to clean the desktop on startup. Please check attached design of my software as well for better idea.

There are two answers:
You can make a program that reads the arguments in the Main method to decide in which mode it should run. If run without arguments (such as when double clicked), it presents the user interface, and if run with some specific argument, like say /run, it does not present the user interface and instead performs the task you want. You don't specify how you get the task to happen at startup, but it will have to run this program with that argument (/run).
The easier way to do this is to just make a new project for the program performing the task - that also lets you start and debug the task directly in Visual Studio.
If you're making a conventional Windows application, you're free to save the settings wherever you'd like. Saving them under a subfolder of the AppData folder in the user's home folder/"profile" is the recommended way. You can use Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData) to get the path to the AppData folder and then make your own folder and files within it.
You can also use the registry.

Passing stream objects / file handles between two processes in .NET

I have a windows service and a desktop application running on the same machine. The app pre-processes some documents and transfers them to a folder where the service can take over. When the app is creating the new file for the service, it keeps a read-only lock on the file while writing. It them releases it so that the service can acquire a new read-only lock (FileStream).
I'd like the app to somehow hand-over this lock to the service without closing it. Is this possible in the managed runtime? If not, is there a way to P/Invoke this behaviour?
The reason this behaviour is desired is so that no other processes can modify or delete the file until both the app and service are done with it.

This is not possible in a managed-only way.
Try to use a simpler approach, such as naming the file with a random name in a temp directory such that no other application will try to open it.
If you insist on passing the handle, you must duplicate the handle into the service process and pass the numeric handle value of that process to the service process. Use OpenProcess, DuplicateHandle and CloseProcess for that.
From http://msdn.microsoft.com/en-us/library/windows/desktop/ms724251(v=vs.85).aspx:
The duplicate handle refers to the same object as the original handle.
Therefore, any changes to the object are reflected through both
handles. For example, if you duplicate a file handle, the current file
position is always the same for both handles. For file handles to have
different file positions, use the CreateFile function to create file
handles that share access to the same file.

Open file on start up

I don't know what it is called but I want to be able to double click on my saved file and then my program should open and load the file. What is this called and how do I do it?
I am using c# wpf and .net 4.0
BR
How about the last 2 fields, what am I supposed to write there?

That is a file association, if you want this to happen on a client machine you need to register your application as the default application for a given extension. This question might be handy.
To actually handle the opening you need to process the arguments that are handed to your application, they will contain the file path. You can get the arguments either in the override of Application.OnStartup (e.Args) or Environment.GetCommandLineArgs.

you need to register the file extension and associate it to your program, either during the setup using certain APIs or from code when program executes the first time.
check these ones:
How to associate a file extension to the current executable in C#
Associate File Extension with Application
personally I do not like the 100% registry approach, there should be some Windows APIs for that and we should let those APIs to work without worrying about the Registry from our side, in my opinion.

How to execute an event of already launched application with file association?

After playing around with a new Windows Form project, I discovered that when you associate a file type with an executable in Windows, you can find the file path of the file that launched the application using args[0] from static void Main(string[] args)
Is it possible to launch an event on your application when you double click a file if your application is already open? (As obviously Main(string[] args) won't be triggered).
Example of an application with behavior I am attempting to replicate:
User Opens GIMP(in Windows)
User opens explorer and right clicks a .png file
User selects open with GIMP
Instead of creating a new application instance of GIMP, GIMP opens the picture in a new window within the instance of GIMP that was already opened.
In this case is GIMP employing multiple applications to accept files "opened" with file association? Or is it possible to do it with a single application "instance".
I'm having trouble with this as most of my searches tend to lead me towards file association as a Windows user (i.e. "How to associate .xls files with excel" articles).

There are a variety of options, but none of them come for free.
Your program's Main() can detect that there is another copy already running and hand the file name off to the already-running copy by some means you determine.
Your program can register as a DDE server and request that subsequent opens be performed via DDE. This is an old-fashioned technique from the 1990's that is generally not recommended for new programs.
You can register a custom drop target for your application, and have the drop target hand the file name to the already-running copy. This mechanism takes the form of a shell extension, and therefore is not suitable for C# due to CLR injection issues.
Given your constraints, option (1) appears to be the best option.

Raymond is right of course, but if you're looking for help with the implmentation of option (1) you should probably look at What is the correct way to create a single instance application? and .NET 4 single application instance and Switch to other instance of same application
You'll notice that detecting the application is pretty easy (use a mutex). Bringing the other application and sending it a filename can be more challenging.
There are three basic solutions presented in the answers to the previously linked questions
Use PostMessage to send a message to 1st instance. This uses HWND_BROADCAST which can have untended consequences.
Use Microsoft.VisualBasic.ApplicationServices.ApplicationBase Of course a reference to VisualBasic gives some C# devs the willies.
Use FindWindow which relies on a Windows Name.
Its also worth noting that if you want the existing application to be in the front you'll need to take special care because setting the foreground can only be given away not taken. See Foreground activation permission is like love: You can't steal it, it has to be given to you and AllowSetForegroundWindow and SetForegroundWindow

Microsoft created this functionality for Visual Basic .Net, and it can be used by C# too.
See the following post for a simple example in C#:
http://www.openwinforms.com/single_instance_application.html

This approach worked best for me: Enforcing Single Instance WPF Applications.
Especially, its solution for passing arguments also works in notify-icon only applications.

Kill process started with System.Diagnostic.Process.Start("FileName")

I am trying to create an app that will perform actions on specific times (much like the Windows Task Scheduler). I am currently using Process.Start() to launch the file (or exe) required by the task.
I am initiating a process by calling a file (an .mp3) and the process starts WMP (since it is the default application). So far so good. Now I want to kill that process. I know that it is normal behavior for the Process.Start(string, string) to return nothing (null in C#) in this case.
So I am asking how can I close WMP when I called it through Process.Start(string, string)??
Edit:
Please note that I am not opening WMP directly with Process.Start() and this is the line with which I run the process:
VB: Me._procs.Add(Process.Start(Me._procInfo))
C#: this._procs.Add(Process.Start(this._procInfo))
_procInfo is a ProcessStartInfo instance. _procInfo.FileName is "C:\route\myFile.mp3". That is why WMP opens. In any case, all of the Start() methods, except for the instance-one which returns a boolean, return nothing (null in C#), because WMP is not the process that was directly created (please note that WMP is run and the song does play).

Process.Start(string,string) returns you a Process resource that you can use to further control the new process.
Process newProcess = Process.Start("param1", "param2");
if (newProcess != null && !newProcess.HasExited)
newProcess.Kill();
The same structure works if you use Process.Start(string), or any other static Process.Start overload.
Process.Start() is a member function and associates a new or reused Process with the Process component identified by this. Behaviour of this method depends on the properties of the Process identified by this.

Don't do it this way.
It's not clear whether the intent of your program is 'Always launch with Windows Media Player' or 'Launch with the registered MP3 player', which might be, say, iTunes.
If you need WMP, use Process.Start with the full path to windows media player.
If you need the registed MP3 player, you can find out the correct exe using the code shown here. Again, start the process with this exe path, passing the MP3 as a parameter.

Two ways:
1-
Process customProc = Process.Start("ExecutablePath", "Argument(s)");
customProc.Kill()
2-
Dim pProcess() As Process = System.Diagnostics.Process.GetProcessesByName("ProcessName")
For Each p As Process In pProcess
p.Kill()
Next

If you are letting the registered windows program to open the file, rather than picking the program you want. Then I advise you do not kill the process.
The reason for this is what say your program does use the default application, but that application is already in use, and contains unsaved data. A user would not be happy for your program to overtake there application with the new file and then kill off the process that was already in use by the user for another purpose. Sure, it might not be in use but you must consider the worst case.
As such, I recommend what has been suggested. use Process.Start() with the full path to the program to be used and the file to be opened.

I tried to open .txt file and the process of my text editor was returned, also I tried .mp3 by WMP and it returned null. So it depends on the application. Do you need to run you mp3 only with WMP? If not, you definitely can create the application which will return the Process object.

proc = Process.Start(filename) should work, but like you say, it returns null instead of a process.
That seems to be inherent to Windows Media Player. Other applications return the process. You can get Windows Media Player's process by specifying the application in the start method.
proc = Process.Start("C:\Program Files\Windows Media Player\wmplayer.exe", filename)
Then you can kill it normally.
proc.Kill()
You will probably need to get the location of the application assiciated with .mp3 files from the registry.