How Start a window with Main() in Console Project.?

Question

I have a console project, but now I need to put an user interface on. So I'm using the 3 tier model (presentation, business, access data).

In my method Main() I call to presentation layer (like app in Window form or Wpf), so, in the presentation layer is the interaction with user through CONSOLE.

Now I add a window called "UserInterface.xaml" in the presentation layer to use instead of console. Because should be with INTERFACE not console.

I have observed that in MainWindow the called is with MainWindow.Show();

But I don't know how to call my "UserInterface.xaml", because has no .Show() method.

This is my method Main:

public static void Main()
{
  MainWindow.Show(); // THIS IS WITH MainWindow.xaml
  UserInterface.???  // THIS IS MY CASE WITH UserInterface.xaml
}

So can somebody tell me how I can call my window from the Main method??

possible duplicate of [Open a Window since Programm class?](http://stackoverflow.com/questions/6073658/open-a-window-since-programm-class) — Cody Gray - on strike, May 21 '11 at 11:44

score 5 · Accepted Answer · answered May 19 '11 at 22:57

5

You definitely got started with the wrong project template. To make a WPF window visible and interactive, you have to follow the rules for a UI thread. Which includes marking the main thread of your app as an STA thread and pumping a message loop. Like this:

class Program {
    [STAThread]
    public static void Main() {
        var app = new Application();
        app.Run(new MainWindow());
    }
}

Beware that Application.Run() is a blocking call, it will not return until the user closes the main window. This is a rather inevitable consequence of the way Windows works.

answered May 19 '11 at 22:57

Hans Passant

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yes, in my case I make this: var app = new UserInterface(); app.Run(new Program()); but not recognizing the .Run – ale May 19 '11 at 23:17
Right, you got that backwards. – Hans Passant May 19 '11 at 23:52
Now I open the window whith [STAThread], but closed again!, why occurs this? what cain I do?? – ale May 20 '11 at 14:18

score 1 · Answer 2 · answered May 19 '11 at 22:44

1

Assuming UserInterface is really a window, this should work:

var window = new UserInterface();
window.Show();

answered May 19 '11 at 22:44

svick

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not recognizing the .Show();... and UserInterface is a Window. – ale May 19 '11 at 23:15
You are going to have to show us more code in that case, because if `UserInterface` were a `Window`, it would have a `Show()` method. How does the source code for `UserInterface` look like? – svick May 19 '11 at 23:28
this is part of the class UserInterface, how you see is a normal window: public class UserInterface: Window { public event EventHandler OptionSelected; public void SetPathCharger2() { InitializeComponent(); } } – ale May 20 '11 at 13:40
@ale, then `UserInterface` *does have* `Show()` method, unless `Window` is not `System.Windows.Window` in your code. – svick May 20 '11 at 14:42

How Start a window with Main() in Console Project.?

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