Open WPF application from console and close console window

Question

I've got an hybrid application with either console or WPF functionality. If the WPF application is started or something is done in console window depends on the arguments at start up. I were able to implement this (there are a lot of examples to find at stackoverflow). Now I want that, if the WPF application is started, that the console window will be closed. But this is shown and if I close it, the WPF application is also closed.

This is my current implementation.

using System;
using System.Windows;

namespace MyNamespace
{

    class Program
    {

        [STAThread]
        static void Main(string[] args)
        {
            string option = args[0];

            switch (option)
            {
                case "WPF":
                    RunApplication();

                    break;
                default:
                    DoSomething();

                    break;
            }
        }

        private static void RunApplication()
        {
            Application app = new Application();
            app.Run(new MainWindow());

            Environment.Exit(0);
        }

        private static void DoSomething()
        {
            // …
        }
    }
}

If I try to start the application in a new Thread the application is directly closed and the WPF window will not be shown.

using System.Threading;
using System.Threading.Tasks;

private static void RunApplication()
{
    new Thread(() => {
        Application app = new Application();
        app.Run(new MainWindow());
    }).Start();

    Environment.Exit(0);
}

I have no idea how I could implement this. Is there a possibility to do this?

@Mat J I've found it in one example. If I remove this line I got an error that the calling `Thread` has to be a STA-`Thread`. — mburm, Mar 09 '19 at 17:48
@Andy I don't think that this will change something. This is the same application which has a different behavior in dependence of the argument but a part is used in both paths. I can devide the application in a console application, WPF application and a shared component. Then I can call the WPF application from the console application and this will be closed. But I would prefer to hold it in one application. — mburm, Mar 09 '19 at 17:53
I could avoid the STA exception after starting the WPF window in a new `Thread`. I have to change the `Thread` start method to `Thread thread = new Thread(...); thread.SetApartmentState(ApartmentState.STA); thread.Start();` The behavior is the same like without a new `Thread`. — mburm, Mar 09 '19 at 18:00

score 0 · Accepted Answer · answered Mar 09 '19 at 18:15

I could find a solution. According of the accepted answer of this post Show/Hide the console window of a C# console application I hide the console window.

using System;
using System.Runtime.InteropServices;
using System.Windows;

namespace DeploymentPreparer
{

    class Program
    {

        [STAThread]
        static void Main(string[] args)
        {
            string option = args[0];

            switch (option)
            {
                case "WPF":
                    RunApplication();

                    break;
                default:
                    DoSomething();

                    break;
            }
        }

        private static void RunApplication()
        {
            ShowWindow(GetConsoleWindow(), SW_HIDE);
            Application app = new Application();
            app.Run(new MainWindow());
        }

        private static void DoSomething()
        {
            // ...
        }

        [DllImport("kernel32.dll")]
        static extern IntPtr GetConsoleWindow();

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

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

Now I have either the console or the WPF window. If the WPF window is shown the console window is hidden.

score 0 · Answer 2 · answered Mar 10 '19 at 18:27

I tried the following method that seems to work: Create a normal Console App. In case of "WPF" argument start the WPF application as a new process. In case of any other argument - call DoSomething()

Example:

using System;
using System.Diagnostics;

namespace ConsoleApp
{
    class Program
    {
        static void Main(string[] args)
        {
            string option = "";
            if (args.Length > 0)
            {
                option = args[0];
            }

            switch (option)
            {
                case "WPF":

                    try
                    {
                        using (Process myProcess = new Process())
                        {
                            myProcess.StartInfo.UseShellExecute = false;

                            // Use correct path to the WPF Application
                            myProcess.StartInfo.FileName = @"C:\Users\Danny\Source\Repo\WpfApp\bin\Debug\WpfApp.exe";
                            myProcess.Start();
                        }
                    }
                    catch (Exception e)
                    {
                        Console.WriteLine(e.Message);
                        Console.WriteLine("Press any key to continue ...");
                        Console.ReadKey();
                    }

                    break;
                default:
                    DoSomething();

                    break;
            }
        }

        private static void DoSomething()
        {
            // …
            Console.WriteLine("Doing Something ...");
            Console.WriteLine("Press any key to continue ...");
            Console.ReadKey();
        }
    }
}

Open WPF application from console and close console window

2 Answers2