Run cmd with parametrs from c# code

Question

Let's say i have button on my form (i use c#)

private void yesButton_Click(object sender, EventArgs e)
{
}

how and what i need to write, so that i can call my ruby cmd file?

for example in cmd i write:

cd c:\ruby\ 
ruby 1.rb 
(here i need to pass some var's) 
root 
123

how can i call this in my c# .net app? also i need to see cmd output on some form component (which is best) ?

also i need to call this in "multi-task", so call this cmd let's say 10 time in ~ on time...

@MichaelPerrenoud i need to start cmd and pass there values inside form app, and then use cmd in form — byCoder
– byCoder, Commented Nov 20, 2013 at 13:39

Mike Perrenoud · Accepted Answer · 2013-11-20 15:01:51Z

1

Alright, so you're question is still real vague, but you can do it like this:

var pathToRbFile = @"C:\Ruby193\script\123.rb";
var arguments = string.Empty; // I don't know what the arguments would be

var info = new ProcessStartInfo(pathToRbFile, arguments);
Process.Start(info);

edited Nov 20, 2013 at 15:01

answered Nov 20, 2013 at 13:42

Mike Perrenoud

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15 Comments

krizz Over a year ago

wouldn't the .rb file be the argument and the first parameter the path to the ruby executable in this case? otherwise this should work.

Mike Perrenoud Over a year ago

@krizz, the .rb file should already be executable by the shell in general. Process.Start leverage the normal shell execution.

Mike Perrenoud Over a year ago

@ValdisAzamaris, I don't know all of that information because I don't know anything about ruby. But as I stated in my answer, the first parameter is the fully qualified path to the .rb file and then the second parameter is a string of arguments. What that path is, and what those arguments are, that's something you'll have to work out.

byCoder Over a year ago

forget about ruby, just imagine that i have rb file in "C:\\Ruby193\\scripts" and there i must execute command ruby 123.rb, how to write code than? also give full code, with creating class instance etc

Mike Perrenoud Over a year ago

@ValdisAzamaris, I've updated my question, but that is all the code. That's all there is to it.

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