Develop Reference

C# - Open a file with the associated program without invoking the command line

Is it possible to open a file with the default program without invoking the command line? I want to run a unit test and have the unit test open the file (PDF) at completion for visual inspection.

Just call Process.Start(filePath).
This will open the file in the user's default program.

I think this should work:
System.Diagnostics.Process.Start(#"c:\file.pdf"); //i.e provide the full path!

Simply use the following syntax:
System.Diagnostics.Process.Start(#"c:\yourfile.txt");

Process process = new System.Diagnostics.Process();
process.EnableRaisingEvents = false;
process.StartInfo.CreateNoWindow = true;
process.StartInfo.FileName = filePath;
string arguments = fileArguments;
process.StartInfo.Arguments = fileArguments;
process.Start();
process.WaitForExit();
This way you can invoke and put the file name in the pdf with parameters/arguments. You can also specify different programs and put it in the path, then the pdf name in the fileArguments. It's up to you.

if you use this code:
System.Diagnostics.Process.Start( "C:\...\...\myfile.pdf" );
the pdf should get opened by the default program associated to the .pdf extension.
is this what you wanted? I would be careful in putting this inside the unit test in case you include those tests in an automated build on the server, which runs with no logged in user, this could be an issue if it fails and if it does not fail, who is there to close Acrobat Reader? :D

Execute OS Command on a file C#

I am trying to execute a OS command through C#. I have the following code taken from this webpage:
//Execute command on file
ProcessStartInfo procStart =
new ProcessStartInfo(#"C:\Users\Me\Desktop\Test\System_Instructions.txt",
"mkdir testDir");
//Redirects output
procStart.RedirectStandardOutput = true;
procStart.UseShellExecute = false;
//No black window
procStart.CreateNoWindow = true;
//Creates a process
System.Diagnostics.Process proc = new System.Diagnostics.Process();
//Set start info
proc.StartInfo = procStart;
//Start
proc.Start();
but when I attempt to run the code I get the following error:
{"The specified executable is not a valid application for this OS platform."}
What am I doing wrong? I have tried this example as well but got the same issue.

The overload of the ProcessStartInfo constructor you are using expects an executable file name and parameters to pass to it - a .txt file is not executable by itself.
It sounds more like you want to execute a batch file with commands within the file. For that check this SO thread: How do I use ProcessStartInfo to run a batch file?

Try setting the USESHELLEXECUTE member to TRUE instead of FALSE.
It worked for me - but I think this has reprocussions for certain users after publishing.

You are trying to execute a TXT file. That's why you get
{"The specified executable is not a valid application for this OS platform."}
Because, well, the specified executable (TXT) is not a valid application for this OS platform.

You would target an executable or other file that has a specified opening application. You're targeting a text file; what you should do is target Notepad, and then supply the path to your text file as an argument:
ProcessStartInfo info = new ProcessStartInfo
{
FileName = "C:\\Windows\System32\\notepad.exe",
Arguments = "C:\\Users\\Me\\Desktop\\Test\\System_Instructions.txt"
}
new Process.Start(info);
Alternatively, if you mean for your text file to be executed, it needs to be made a .bat file.

You are trying to execute this:
C:\Users\Me\Desktop\Test\System_Instructions.txt mkdir testDir
The shell has no clue how to "execute" a text file so the command fails.
If you want to execute this text file as a batch file, change file extension to .bat so the system understands it's a batch file, and then set UseShellExecute so it does the default action for it (= runs it, in case of a batch file).

If you want to open up the file in Notepad, use:
ProcessStartInfo procStart =
new ProcessStartInfo("notepad.exe", #"C:\Users\Me\Desktop\Test\System_Instructions.txt");
If you want to write into the file :
//In case the directory doesn't exist
Directory.CreateDirectory(#"C:\Users\Me\Desktop\Test\);
using (var file = File.CreateText(#"C:\Users\Me\Desktop\Test\System_Instructions.txt"))
{
file.WriteLine("mkdir testDir");
}
If you have commands in the text file that you want to execute, just rename it to .bat and it should work (and presumably the contents do something with "mkdir testDir" as a parameter?)

What are you trying to accomplish?
Create a directory? Use the "System.IO.Directory.CreateDirectory" method.
Open a .txt file with associated program? Use ProcessStartInfo(#".\filename.txt") with UseShellExecute set to true. This will cause the associated program for that file type to be executed, which might not be notepad.txt.

C# Open File With Associated Application passing arguments

I am trying to launch the default application registered for an extension specifying an additional argument:
ProcessStartInfo p = new ProcessStartInfo();
p.Arguments = "myargument";
p.FileName = "file.ext";
Process.Start(p);
The application starts correctly opening the specified file.
The problem is that it is getting just one parameter (the name of the file), totally ignoring the additional "Arguments".
Is it possible to do what I want?
Am I doing something wrong?
Thanks in advance for any help,
Paolo

I believe this is expected. Behind the scenes, Windows is finding the default application in the registry and creating a new process and passing your file name to it. I get the same behavior if I go to a command prompt and type "filename.ext argument", that my arguments are not passed to the application.
What you probably need to do is find the default application yourself by looking in the registry. Then you can start that process with arguments, instead of trying to start by filetype association. There is an answer here on how to find the default application in the registry:
Finding the default application for opening a particular file type on Windows

what exactly is your "argument", does it have spaces, backslash, etc?
Process process = new Process();
process.StartInfo.FileName = #"C:\process.exe";
process.StartInfo.Arguments = #"-r -d something else";
process.StartInfo.CreateNoWindow = false;
process.StartInfo.UseShellExecute = false;
process.Start();
Is there any reason why you cant start the app, then use the extension and arguments in your arguments?

I think an easier method is using the cmd command
void LaunchAssociatedProgram(string filename) {
Process.Start( #"cmd.exe", "/C start "+ filename );
}
EDIT:
I don't know if it works with arguments, but it is what I was looking for to launch an associated program...

Open a file with Notepad in C#

How I open a file in c#? I don't mean reading it by textreader and readline(). I mean open it as an independent file in notepad.

You need System.Diagnostics.Process.Start().
The simplest example:
Process.Start("notepad.exe", fileName);
More Generic Approach:
Process.Start(fileName);
The second approach is probably a better practice as this will cause the windows Shell to open up your file with it's associated editor. Additionally, if the file specified does not have an association, it'll use the Open With... dialog from windows.
Note to those in the comments, thankyou for your input. My quick n' dirty answer was slightly off, i've updated the answer to reflect the correct way.

You are not providing a lot of information,
but assuming you want to open just any file on your computer
with the application that is specified for the default handler for that filetype,
you can use something like this:
var fileToOpen = "SomeFilePathHere";
var process = new Process();
process.StartInfo = new ProcessStartInfo()
{
UseShellExecute = true,
FileName = fileToOpen
};
process.Start();
process.WaitForExit();
The UseShellExecute parameter tells Windows to use the default program for the type of file you are opening.
The WaitForExit will cause your application to wait until the application you luanched has been closed.

this will open the file with the default windows program (notepad if you haven't changed it);
Process.Start(#"c:\myfile.txt")

System.Diagnostics.Process.Start( "notepad.exe", "text.txt");

You can use Process.Start, calling notepad.exe with the file as a parameter.
Process.Start(#"notepad.exe", pathToFile);

Use System.Diagnostics.Process to launch an instance of Notepad.exe.

how to create command application like run command for vista

how to create application like window run command using C#. When i insert any command (for example: ipconfig) , this return result (for example: 192.168.1.1) on the textbox.
how to get windows command list?
how to get command result?
how to get installed application list on the machine?

(1) The command list will most likely come from whatever executables are found in your %PATH%. You can figure out your list by finding all .exe/.bat/other executable files in every folder specified by %PATH%. You may not even need to know which apps are available, the Process.Start method will find them for you. (see below)
(2) You can run a command-line tool programmatically using:
System.Diagnostics.Process.Start("notepad.exe"); // located using %PATH%
To capture the output, you have to redirect it like this:
System.Diagnostics.ProcessStartInfo psi =
new System.Diagnostics.ProcessStartInfo(#"ipconfig");
psi.RedirectStandardOutput = true;
psi.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
psi.UseShellExecute = false;
System.Diagnostics.Process myProcess;
myProcess = System.Diagnostics.Process.Start(psi);
System.IO.StreamReader myOutput = myProcess.StandardOutput; // Capture output
myProcess.WaitForExit(2000);
if (myProcess.HasExited)
{
string output = myOutput.ReadToEnd();
Console.WriteLine(output);
}
(3) Probably the same answer as 1

Create a Windows Forms application using the wizard. Draw a text box and a button. Add a Click handler to the button which takes the contents of the text box and launches a process. Use the Process class. That class also has a StandardOutput property that you can read the output from so you can put it into the text box.
You may find that to use many Command Prompt commands, you need to type CMD /C in front, because they aren't separate programs from the command interpreter.
As for discovering a list of commands, that's not generally possible. A command is just a program (or a feature of the CMD command interpreter). You could search the hard drive for .exe files, but then many of them wouldn't be suitable as "commands".