Process.start: how to get the output?

1: Synchronous example

static void runCommand() {
Process process = new Process();
process.StartInfo.FileName = "cmd.exe";
process.StartInfo.Arguments = "/c DIR"; // Note the /c command (*)
process.StartInfo.UseShellExecute = false;
process.StartInfo.RedirectStandardOutput = true;
process.StartInfo.RedirectStandardError = true;
process.Start();
//* Read the output (or the error)
string output = process.StandardOutput.ReadToEnd();
Console.WriteLine(output);
string err = process.StandardError.ReadToEnd();
Console.WriteLine(err);
process.WaitForExit();
}

Note that it's better to process both output and errors: they must be handled separately.

(*) For some commands (here

StartInfo.Arguments

) you must add the

/c

directive, otherwise the process freezes in the

WaitForExit()

.

2: Asynchronous example

static void runCommand() {
//* Create your Process
Process process = new Process();
process.StartInfo.FileName = "cmd.exe";
process.StartInfo.Arguments = "/c DIR";
process.StartInfo.UseShellExecute = false;
process.StartInfo.RedirectStandardOutput = true;
process.StartInfo.RedirectStandardError = true;
//* Set your output and error (asynchronous) handlers
process.OutputDataReceived += new DataReceivedEventHandler(OutputHandler);
process.ErrorDataReceived += new DataReceivedEventHandler(OutputHandler);
//* Start process and handlers
process.Start();
process.BeginOutputReadLine();
process.BeginErrorReadLine();
process.WaitForExit();
}
static void OutputHandler(object sendingProcess, DataReceivedEventArgs outLine) {
//* Do your stuff with the output (write to console/log/StringBuilder)
Console.WriteLine(outLine.Data);
}

If you don't need to do complicate operations with the output, you can bypass the OutputHandler method, just adding the handlers directly inline:

//* Set your output and error (asynchronous) handlers
process.OutputDataReceived += (s, e) => Console.WriteLine(e.Data);
process.ErrorDataReceived += (s, e) => Console.WriteLine(e.Data);