Running a Linux Console Command in C#

Question

I am using the following code to run a Linux console command via Mono in a C# application:

ProcessStartInfo procStartInfo = new ProcessStartInfo("/bin/bash", "-c ls");
procStartInfo.RedirectStandardOutput = true;
procStartInfo.UseShellExecute = false;
procStartInfo.CreateNoWindow = true;

System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.StartInfo = procStartInfo;
proc.Start();

String result = proc.StandardOutput.ReadToEnd();

This works as expected. But, if i give the command as "-c ls -l" or "-c ls /path" I still get the output with the -l and path ignored.

What syntax should I use in using multiple switches for a command?

You could try using ProcessStartInfo.Arguments to see if the alternative method works? Also do you need the /bin/bash? can you not just run 'ls' direct? — cjb110
– cjb110, Commented Jul 23, 2014 at 12:49
@cjb110 no it doesn't work. Yes you do have to set /bin/bash as the filename or it cannot find the bash executable on its own. — ravindu1024
– ravindu1024, Commented Jul 23, 2014 at 13:04
Maybe try RedirectStandardInput and send the commands. I don't know the exact code but I do know you can do this to send input to the process. Here is a sample: msdn.microsoft.com/en-us/library/… — NewfrontSolutions
– NewfrontSolutions, Commented Jul 23, 2014 at 13:27
Is there a solution to this problem? I am having the same one. — Vaibhav Aggarwal
– Vaibhav Aggarwal, Commented Jul 28, 2014 at 23:28

Fab · Accepted Answer · 2015-10-04 04:11:46Z

3

You forgot to quote the command.

Did you try the following on the bash prompt ?

bash -c ls -l

I strongly suggest to read the man bash. And also the getopt manual as it's what bash use to parse its parameters.

It has exactly the same behavior as bash -c ls Why? Because you have to tell bash that ls -l is the full argument of -c, otherwise -l is treated like an argument of bash. Either bash -c 'ls -l' or bash -c "ls -l" will do what you expect. You have to add quotes like this:

ProcessStartInfo procStartInfo = new ProcessStartInfo("/bin/bash", "-c 'ls -l'");

edited Oct 4, 2015 at 4:11

answered Sep 19, 2015 at 11:59

Fab

14.9k5 gold badges54 silver badges73 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user1709408 Over a year ago

bash -c 'ls -l' is almost the same as bash -c "ls -l" but doesn't require escaping in C# string

Collectives™ on Stack Overflow

Running a Linux Console Command in C#

1 Answer 1

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related