CODEWITHSUNDEEP
c#consolewinforms
Dervin Thunk
Dervin Thunk

Reputation: 20119

Windows form from console

I would like to spawn a Windows form from the console using C#. Roughly like display does in Linux, and modify its contents, etc. Is that possible?

Upvotes: 5

Views: 14985

Answers (4)

Philip Wallace
Philip Wallace

Reputation: 8015

The common answer:

[STAThread]
static void Main()
{    
   Application.Run(new MyForm());
}

Alternatives (taken from here) if, for example - you want to launch a form from a thread other than that of the main application:

Thread t = new Thread(new ThreadStart(StartNewStaThread)); 

// Make sure to set the apartment state BEFORE starting the thread. 
t.ApartmentState = ApartmentState.STA; 
t.Start(); 

private void StartNewStaThread() { 
    Application.Run(new Form1()); 
}

.

Thread t = new Thread(new ThreadStart(StartNewStaThread)); 
t.Start();

[STAThread]
private void StartNewStaThread() { 
    Application.Run(new Form1()); 
}

Upvotes: 4

RRUZ
RRUZ

Reputation: 136391

You can try this

using System.Windows.Forms;

[STAThread]
static void Main() 
{
    Application.EnableVisualStyles();
    Application.Run(new MyForm()); 
}

Bye.

Upvotes: 1

Matthew Whited
Matthew Whited

Reputation: 22443

You should be able to add a reference for System.Windows.Forms and then be good to go. You may also have to apply the STAThreadAttribute to the entry point of your application. 

using System.Windows.Forms;

class Program
{
    [STAThread]
    static void Main(string[] args)
    {
        MessageBox.Show("hello");
    }
}

... more complex ...

using System.Windows.Forms;

class Program
{
    [STAThread]
    static void Main(string[] args)
    {
        var frm = new Form();
        frm.Name = "Hello";
        var lb = new Label();
        lb.Text = "Hello World!!!";
        frm.Controls.Add(lb);
        frm.ShowDialog();
    }
}

Upvotes: 6

Yuriy Faktorovich
Yuriy Faktorovich

Reputation: 68687

Yes, you can initialize a form in the Console. Add a reference to System.Windows.Forms and use the following sample code:

System.Windows.Forms.Form f = new System.Windows.Forms.Form(); 
f.ShowDialog();

Upvotes: 4

Related Questions