How to run a C# console application with the conso

问题:

Is there a way to hide the console window when executing a console application?

I am currently using a Windows Forms application to start a console process, but I don't want the console window to be displayed while the task is running.

回答1:

If you are using the ProcessStartInfo class you can set the window style to hidden:

System.Diagnostics.ProcessStartInfo start =
      new System.Diagnostics.ProcessStartInfo();
start.FileName = dir + @"\Myprocesstostart.exe";
start.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;

回答2:

If you wrote the console application you can make it hidden by default.

Create a new console app then then change the "Output Type" type to "Windows Application" (done in the project properties)

回答3:

If you are using Process Class then you can write

yourprocess.StartInfo.UseShellExecute = false;
yourprocess.StartInfo.CreateNoWindow = true;

before yourprocess.start(); and process will be hidden

回答4:

Simple answer is that: Go to your console app's properties(project's properties).In the "Application" tab, just change the "Output type" to "Windows Application". That's all.

回答5:

You can use the FreeConsole API to detach the console from the process :

[DllImport("kernel32.dll")]
static extern bool FreeConsole();

(of course this is applicable only if you have access to the console application's source code)

回答6:

If you're interested in the output, you can use this function:

private static string ExecCommand(string filename, string arguments)
{
    Process process = new Process();
    ProcessStartInfo psi = new ProcessStartInfo(filename);
    psi.Arguments = arguments;
    psi.CreateNoWindow = true;
    psi.RedirectStandardOutput = true;
    psi.RedirectStandardError = true;
    psi.UseShellExecute = false;
    process.StartInfo = psi;

    StringBuilder output = new StringBuilder();
    process.OutputDataReceived += (sender, e) => { output.AppendLine(e.Data); };
    process.ErrorDataReceived += (sender, e) => { output.AppendLine(e.Data); };

    // run the process
    process.Start();

    // start reading output to events
    process.BeginOutputReadLine();
    process.BeginErrorReadLine();

    // wait for process to exit
    process.WaitForExit();

    if (process.ExitCode != 0)
        throw new Exception("Command " + psi.FileName + " returned exit code " + process.ExitCode);

    return output.ToString();
}

It runs the given command line program, waits for it to finish and returns the output as string.

回答7:

If you're creating a program that doesn't require user input you could always just create it as a service. A service won't show any kind of UI.

回答8:

I know I'm not answering exactly what you want, but I am wondering if you're asking the right question.

Why don't you use either:

1. windows service
2. create a new thread and run your process on that

Those sound like better options if all you want is to run a process.

回答9:

Add this to your class to import the DLL file:

[DllImport("user32.dll")]
    static extern bool ShowWindow(IntPtr hWnd, int nCmdShow);

    const int SW_HIDE = 0;
    const int SW_SHOW = 5;

And then if you want to hide it use this command:

var handle = GetConsoleWindow();
ShowWindow(handle, SW_HIDE);

And if you want to show the console:

var handle = GetConsoleWindow();
ShowWindow(handle, SW_SHOW);

回答10:

Based on Adam Markowitz's answer above, following worked for me:

process = new Process();
process.StartInfo = new ProcessStartInfo("cmd.exe", "/k \"" + CmdFilePath + "\"");
process.StartInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
//process.StartInfo.UseShellExecute = false;
//process.StartInfo.CreateNoWindow = true;
process.Start();

回答11:

I've got a general solution to share:

using System;
using System.Runtime.InteropServices;

namespace WhateverNamepaceYouAreUsing
{
    class Magician
    {
    [DllImport("kernel32.dll")]
        static extern IntPtr GetConsoleWindow();

        [DllImport("user32.dll")]
        static extern bool ShowWindow(IntPtr hWnd, int nCmdShow);

        const int HIDE = 0;
        const int SHOW = 5;

        public static void DisappearConsole()
        {
            ShowWindow(GetConsoleWindow(), HIDE);
        }
    }
}

Just include this class in your project, and call Magician.DisappearConsole();.

A console will flash when you start the program by clicking on it. When executing from the command prompt, the command prompt disappears very shortly after execution.

I do this for a Discord Bot that runs forever in the background of my computer as an invisible process. It was easier than getting TopShelf to work for me. A couple TopShelf tutorials failed me before I wrote this with some help from code I found elsewhere. ;P

I also tried simply changing the settings in Visual Studio > Project > Properties > Application to launch as a Windows Application instead of a Console Application, and something about my project prevented this from hiding my console - perhaps because DSharpPlus demands to launch a console on startup. I don't know. Whatever the reason, this class allows me to easily kill the console after it pops up.

Hope this Magician helps somebody. ;)

回答12:

Just write

ProcessStartInfo psi= new ProcessStartInfo("cmd.exe");
......

psi.CreateNoWindow = true;

回答13:

Although as other answers here have said you can change the "Output type" to "Windows Application", please be aware that this will mean that you cannot use Console.In as it will become a NullStreamReader.

Console.Out and Console.Error seem to still work fine however.