Hi ,
I trying execute the like
ProcessStartInfo proc = new ProcessStartInfo();
proc.FileName ="talks.sh";
proc.WorkingDirectory = FilePath;
proc.UseShellExecute = true;
Process.Start(proc);
But here Code is executed perfectly. But here my file “talks.sh” is not executed in the Linux operating system.
using System.Diagnostics; // or import System.Diagnostics;
// .. stuff
ProcessStartInfo proc = new ProcessStartInfo();
proc.FileName = "xdg-open";
proc.WorkingDirectory = "somepath";
proc.Arguments = "talk.sh";
proc.WindowStyle = ProcessWindowStyle.Minimized;
proc.CreateNoWindow = true;
Process.Start(proc);
ProcessStartInfo proc = new ProcessStartInfo();
proc.FileName = “xdg-open”;
proc.WorkingDirectory = “somepath”;
proc.Arguments = “talk.sh”;
proc.WindowStyle = ProcessWindowStyle.Minimized;
proc.CreateNoWindow = true;
Process.Start(proc);
I tryied above code. there is “talk.sh” file is opened. But it is not executing.
if “talk.sh” is excute , there will be create one more file. file name “create.txt” .