I have a form with 3 controls:
The best solution I have found is:
private void Redirect(StreamReader input, TextBox output)
{
new Thread(a =>
{
var buffer = new char[1];
while (input.Read(buffer, 0, 1) > 0)
{
output.Dispatcher.Invoke(new Action(delegate
{
output.Text += new string(buffer);
}));
};
}).Start();
}
private void Window_Loaded(object sender, RoutedEventArgs e)
{
process = new Process
{
StartInfo = new ProcessStartInfo
{
CreateNoWindow = true,
FileName = "app.exe",
RedirectStandardError = true,
RedirectStandardOutput = true,
UseShellExecute = false,
}
};
if (process.Start())
{
Redirect(process.StandardError, textBox1);
Redirect(process.StandardOutput, textBox1);
}
}
I've used code something like this:
public static void Run(string fileName, string arguments, out string standardOutput, out string standardError, out int exitCode)
{
Process fileProcess = new Process();
fileProcess.StartInfo = new ProcessStartInfo
{
FileName = fileName,
Arguments = arguments,
RedirectStandardError = true,
RedirectStandardOutput = true,
UseShellExecute = false,
WindowStyle = ProcessWindowStyle.Hidden,
CreateNoWindow = true,
};
bool started = fileProcess.Start();
if (started)
{
fileProcess.WaitForExit();
}
else
{
throw new Exception("Couldn't start");
}
standardOutput = fileProcess.StandardOutput.ReadToEnd();
standardError = fileProcess.StandardError.ReadToEnd();
exitCode = fileProcess.ExitCode;
}
But it's not interactive. But if the app is interactive, it'll take a lot more code anyway.
If you want something interactive, I got this code to work (yours modified, details on modifications below)
private void InitializeInterpreter()
{
InterProc.StartInfo.UseShellExecute = false;
InterProc.StartInfo.FileName = "Echoer.exe";
InterProc.StartInfo.RedirectStandardInput = true;
InterProc.StartInfo.RedirectStandardOutput = true;
InterProc.StartInfo.RedirectStandardError = true;
InterProc.StartInfo.CreateNoWindow = true;
InterProc.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
InterProc.OutputDataReceived += new DataReceivedEventHandler(InterProcOutputHandler);
bool started = InterProc.Start();
InterProc.BeginOutputReadLine();
}
private void AppendTextInBox(TextBox box, string text)
{
if (this.InvokeRequired)
{
this.Invoke((Action<TextBox, string>)AppendTextInBox, OutputTextBox, text);
}
else
{
box.Text += text;
}
}
private void InterProcOutputHandler(object sendingProcess, DataReceivedEventArgs outLine)
{
AppendTextInBox(OutputTextBox, outLine.Data + Environment.NewLine);
}
private void Enterbutton_Click(object sender, EventArgs e)
{
InterProc.StandardInput.WriteLine(CommandTextBox.Text);
}
So, I moved the BeginOutputReadLine to just after the process is started. That ensures it's really only called once. I also did an invoke required to clean up thread calls. Hopefully this should work for you.
Where are you calling StandardOutput.ReadToEnd()
? I once had a similar problem because I was calling Process.WaitForExit()
before StandardOutput.ReadToEnd()
. I had a large amount of input, and the output buffer was full before completion and my process was blocked.
You must call StandardOutput.ReadToEnd()
before Process.WaitForExit()
.