Reputation: 141
I'm starting cmd as a process in C# and I want to pass a file path in the argument. How to do it?
Process CommandStart = new Process();
CommandStart.StartInfo.UseShellExecute = true;
CommandStart.StartInfo.RedirectStandardOutput = false;
CommandStart.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
CommandStart.StartInfo.FileName = "cmd";
CommandStart.StartInfo.Arguments = "(here i want to put a file path to a executable) -arg bla -anotherArg blabla < (and here I want to put another file path)";
CommandStart.Start();
CommandStart.WaitForExit();
CommandStart.Close();
EDIT:
Process MySQLDump = new Process();
MySQLDump.StartInfo.UseShellExecute = true;
MySQLDump.StartInfo.RedirectStandardOutput = false;
MySQLDump.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
MySQLDump.StartInfo.FileName = "cmd";
MySQLDump.StartInfo.Arguments = "/c \"\"" + MySQLDumpExecutablePath + "\" -u " + SQLActions.MySQLUser + " -p" + SQLActions.MySQLPassword + " -h " + SQLActions.MySQLServer + " --port=" + SQLActions.MySQLPort + " " + SQLActions.MySQLDatabase + " \" > " + SQLActions.MySQLDatabase + "_date_" + date + ".sql";
MySQLDump.Start();
MySQLDump.WaitForExit();
MySQLDump.Close();
Upvotes: 0
Views: 3499
Reputation: 69372
You need to put the file path in double quotes and use a verbatim string literal (@
) as SLaks mentioned.
CommandStart.StartInfo.Arguments = @"""C:\MyPath\file.exe"" -arg bla -anotherArg";
Example with an OpenFileDialog
using(OpenFileDialog ofd = new OpenFileDialog())
{
if (ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
string filePath = "\"" + ofd.FileName + "\"";
//..set up process..
CommandStart.StartInfo.Arguments = filePath + " -arg bla -anotherArg";
}
}
Update to comment
You can format your string using String.Format
.
string finalPath = String.Format("\"{0}{1}_date_{2}.sql\"", AppDomain.CurrentDomain.BaseDirectory, SQLActions.MySQLDatabase, date);
Then pass finalPath
into the arguments.
Upvotes: 1