Every time I click the No button (here button2_click), the program opens itself again, and I cannot figure out why.
private void button2_Click(object sender, EventArgs e)
{
Process game = new Process();
game.StartInfo.FileName = "ShovelShovel.exe";
game.Start();
}
I figured it would launch the game and close, but if I click the button, it just launches my program again. If I click No on that form, then I get a huge list of unhandled exceptions.
Whether I do this.Close(); or Application.Exit(); the program still opens itself.
Can anyone tell me why it would open again? I haven't altered Program.cs in anyway.
Thank you very much.
Try this: (assuming that the program you are working from is "ShovelShovel Settings.exe" and is the launcher for "ShovelShovel.exe")
//launch game
ProcessStartInfo startInfo = new ProcessStartInfo();
startInfo.FileName = "ShovelShovel.exe";
Process p = Process.Start(startInfo);
//close
this.Close();
Or even just
Process.Start("ShovelShovel.exe");
If this is still starting the same application, you either have the wrong path, or the exe's are the same.
To double check, why dont you go into the directory and execute "ShovelShovel.exe" manually from WindowsExplorer.
Related
I'm rather new to C# so I'm not sure if this is even possible...
What I'm trying to do is use an application form to launch other programs via Process, such as Notepad, and then close my original application while Notepad is still running.
Process notePad = new Process();
notePad.StartInfo.FileName = "notepad.exe";
notePad.Start();
Application.Exit();
From what I've found, I have to close/kill the Notepad process in order to close the launching application. I'd like Notepad to remain open, however.
Is there any way to achieve this via C#?
Try this?
using(Process notePad = Process.Start("notepad.exe","")) { }
Application.Exit();
if you want notepad still open after close the form use this code
private void button1_Click(object sender, EventArgs e)
{
Thread notepad_opener = new Thread(opener);
notepad_opener.Start();
}
public void opener()
{
Process.Start("notepad.exe");
Application.Exit();
}
i try this and notepad still open :D
You need to set the UseShellExecute property to True; see this page
You can save the Id of Process you have started:
int id = notePad.Id;
Then Kill it when you want:
Process.GetProcessById(id).Kill();
I am Working in Visual Studio 2008 Winforms Application project in Windows 7 (32 bit).I am doing the project in C#.
I have placed some buttons in a tab and added actions for that once it is clicked. While clicking the button am just running a .exe file in its action part.
My problem is that, i opened a window by clicking one button(so the .exe file is running), now while am clicking the button again it is opening same window again irrespective of checking that it is open or not. I want to solve this issue,as when a window is opened it must not open again on another click on same button. How to solve this issue. ?
Please help....
Thanks in advance..
You could check if the process is already running, when re-clicking the button:
private void btnStartExecutable_Click(object sender, EventArgs e)
{
Process[] processName = Process.GetProcessesByName("InsertProcessNameHere");
if (pname.Length == 0)
{
MessageBox.Show("Application isn't running yet.");
//Start application here
Process.Start("InsertProcessNameHere");
}
else
{
MessageBox.Show("Application is already running.");
//Don't start application, since it has been started already
}
}
You can try this:
bool processExited = true;
private void button1_Click(object sender, EventArgs e)
{
if (processExited)
{
Process process = new Process();
process.EnableRaisingEvents = true;
process.Exited += MyProcessExited;
process.StartInfo = new ProcessStartInfo();
process.StartInfo.FileName = "notepad.exe";
process.Start();
processExited = false;
}
else
{
MessageBox.Show("Still running");
}
}
void MyProcessExited(object sender, EventArgs e)
{
processExited = true;
}
The right answer here IMHO is that unless the two application shares a common resource or can talk to each other through some channel, there is no safe and efficient way to achieve what you want. Since the process is external, it could already be running before your calling app starts, or even while it's already running. You won't be able to tell if the process has been started from your app or not.
By the time I'm writing this your question does not yet state if you are in liberty to modify the external app you are calling. If you are however, using a Mutex would be a quick and easy way to solve your problem.
In your external app, whenever you want to make the other app aware of whatever condition you want (be it that the process is running or that a specific window is opened), have a Mutex instance created like this:
var mutex = new Threading.Mutex(true, "mutex unique identifier");
And in your calling app, try to create a Mutex instance with the same identifier:
bool alreadyExists;
var mutex = new Threading.Mutex(false, "mutex unique identifier", out alreadyExists);
Here the alreadyExists variable will tell you whether or not the external process is running or not. This is much safer than trying to identify it via its name, as other processes could have the same or a new version could be of a different name. Of course, the mutex identifier must be as unique as possible (like a Guid), otherwise you may encounter the same problem. ;)
Whenever you feel like the mutex must be released (at external app level), release it:
mutex.ReleaseMutex();
Note that if the process ends the mutex will be automatically released by the OS.
If the external app isn't a .NET based app, you can still create a mutex with Win32 API functions.
Thanks for the support.. I got the answer like this..
1) Creating an event'Exit' for the process in function button click
2) Define a function for the exit event where you set a flag
3) Check the flag is set or not everytime while opening the process in the function button click
Event for Exit: 'P' is the name of process:
p.Exited += new EventHandler(p_Exited);
p_Exited will be the function name where we will set the flag.
Thanks all...
If you know the name of the process that gets started or the path the .exe is run from you can use the Process class to check to see if it is currently running.
http://msdn.microsoft.com/en-us/library/system.diagnostics.process(v=vs.110).aspx
I am writing a client for my gaming community and one of the functions of this client is to launch a game via the client with parameters that will enable our community mod pack on launch.
When I press the button, the game begins to launch and as soon as the program opens (the icon pops up in the task bar), it closes instantly.
Is there something I am missing that is needed to keep the launched exe running?
Here is my code:
private void btnLaunchGame_Click(object sender, EventArgs e)
{
string armaPath = gameDir+"/Expansion/beta/";
string filename = Path.Combine(armaPath, "arma2oa.exe");
string launchParams = "-noSplash -noFilePatching -showScriptErrors \"-name=Meta\" \"-mod=I:/Steam/steamapps/common/Arma 2;expansion;expansion/beta;expansion/beta/expansion;servermods/#HC_DAYZ;servermods/#HC_WEAPONS;servermods/#HC_EXTRAS;servermods/#HC_ACE\"";
System.Diagnostics.Process.Start(filename, launchParams);
}//close Game Launch
Any ideas is appreciated!
I have a .bat file that will execute the game flawlessly with the launch args listed below, this could possibly help pinpoint the cause of my problem:
http://puu.sh/5CGKk.png (couldn't get code to paste in a readable format).
Try using Process:
Process process = new Process();
process.StartInfo.FileName = "arma2oa.exe";
process.StartInfo.Arguments = "-noSplash -noFilePatching -showScriptErrors \"-name=Meta\" \"-mod=I:/Steam/steamapps/common/Arma 2;expansion;expansion/beta;expansion/beta/expansion;servermods/#HC_DAYZ;servermods/#HC_WEAPONS;servermods/#HC_EXTRAS;servermods/#HC_ACE\"";
process.StartInfo.WorkingDirectory = gameDir + "/Expansion/beta/";
process.Start();
It may be what exe require working directory to be set. Or it will crash, unable to load resources.
If that doesn't works, then perhaps you need to add
process.WaitForInputIdle();
before exiting function running process. I don't know why, but running Acrobat Reader without this wait may sometimes cause a wierd effect: Acrobat is running, but the document, passed via arguments, is not shown. Perhaps something to do with Garbage collector or Process itself.
Try following
using (Process process = new Process())
{
ProcessStartInfo startInfo = new ProcessStartInfo("C:\Program Files\Arma2oa\Arma2oa.exe");
startInfo.Arguments = "-noSplash -noFilePatching -showScriptErrors \"-name=Meta\" \"-mod=I:/Steam/steamapps/common/Arma 2;expansion;expansion/beta;expansion/beta/expansion;servermods/#HC_DAYZ;servermods/#HC_WEAPONS;servermods/#HC_EXTRAS;servermods/#HC_ACE\"";
process.StartInfo = startInfo;
process.Start();
}
In order to start another instance of my program I did something like:
private void button1_Click(object sender, EventArgs e)
{
Process p = new Process();
p.StartInfo.FileName = Application.ExecutablePath;
p.StartInfo.WindowStyle = ProcessWindowStyle.Normal;
p.Start();
}
And found that stopping the debugger didn't stop the new window, only the first (-launching) window.
How do I programmatically get the new process to be "under" VS?
That Process that you get back has a handle to the running process. You could keep a hold of that in a member variable, rather than a local variable, and on form closing, kill the process.
http://msdn.microsoft.com/en-us/library/e8zac0ca.aspx
You can Change the Start Action for Application Debugging
Right click on your project
Properties
Debug
Start external program
And set the program you want to launch.
If you want to attach to an another instance programmatically, a duplicate question can be found here:
How do I attach a process to the debugger in Visual Studio?
Wich refer this article:
Attaching to a Process using VS.NET Automation Model
Since you're starting your own program a second time, you know it is a GUI. You can keep the Process reference around and call CloseMainWindow (or Kill) on each of them in your FormClosing event handler:
private List<Process> children = new List<Process>();
private void button1_Click(object sender, EventArgs e)
{
Process p = new Process();
p.StartInfo.FileName = Application.ExecutablePath;
p.StartInfo.WindowStyle = ProcessWindowStyle.Normal;
p.Start();
children.Add(p);
}
private Form1_FormClosing(object sender, FormClosingEventArgs e)
{
foreach (Process p in this.children)
{
// posts WM_CLOSE to the main handle of the process
// which allows a graceful exit, as if the user clicked [X]
p.CloseMainWindow();
// p.Kill(); // less graceful, just kill
}
}
Debug -> Attach to Process and select your process from the list.
If by "under VS" you mean having Visual Studio be able to debug the external process you may want to consider the "Attach to Process" strategy.
http://msdn.microsoft.com/en-us/library/c6wf8e4z.aspx
I am working on a Windows application. I have created a help file (.chm) using a third party tool and I call it from the Windows application. The help file is invoked successfully when I click on Help menu item or press F1 for the application.
The problem I am facing is that if I click on the menu item again or press F1 (while the application is open and I have not closed the previous help file) it opens another instance of the help file.
Let me be clear by giving an example: I want it to be just like the "Solitaire" game. No matter how many times you press F1 or the Contents menu item it shows only one help window.
I want to do same thing like "Solitaire". I don't want multiple instances to be created.
I hope you understood my problem. Please let me know if my query is wrong.
I have posted my code below.
ProcessStartInfo info;
Process exeprocess;
The below code is in Help menuitem click event.
private void mnuContents_Click(object sender, EventArgs e)
{
string ApplicationPath=ConfigurationSettings.AppSettings["HelpFile"].ToString();
info = new ProcessStartInfo(ApplicationPath);
//Process.Start(info);
exeprocess = Process.Start(info);
}
One solution is:
Have your application create a system-wide resource (the example below uses a Win32 mutex)
Check the resource before you spawn the .chm (I imagine you're probably using ShellExec or some variant to spawn the help file.
Here's example code (in C++/Win32 code):
http://support.microsoft.com/kb/243953
Another, different approach is to see if any currently running processes match the one you would spawn. Here's example code for this approach:
http://www.dotnetperls.com/single-instance-windows-form
You have a Process object, so you should probably store it somewhere and check if it is still active the next time the help command is invoked. You could use Process.HasExited for that purpose. If it has exited, clean up the Process object by calling Dispose() and then launch a new instance, storing it away again. Repeat as needed.
Ok this is your block of code to start the CHM viewer:
private void mnuContents_Click(object sender, EventArgs e)
{
string ApplicationPath=ConfigurationSettings.AppSettings["HelpFile"].ToString();
info = new ProcessStartInfo(ApplicationPath);
//Process.Start(info);
exeprocess = Process.Start(info);
}
in exeprocess there is a property called Id. You need to keep track of that Id for the next time the user presses F1 or the menu key.
You need to do a check like
private void mnuContents_Click(object sender, EventArgs e)
{
if (Process.GetProcessById(self.previousId) != null) {
string ApplicationPath=ConfigurationSettings.AppSettings["HelpFile"].ToString();
info = new ProcessStartInfo(ApplicationPath);
//Process.Start(info);
exeprocess = Process.Start(info);
self.previousId = exeprocess.Id;
}
}
Something like that would work. If you want to get fancy, you can bring the already-running process to the foreground as well.