如何从另一个类（ApplicationContext）显示已实例化（运行）的Form1类

Question

I have a class for Application Context and in it, among other things i have some methods to send a Form1 class to system-tray, my problem is how to show already instantiated class From1 while still running in system-tray, example:

class MyApplicationContext : ApplicationContext
{
    private NotifyIcon TrayIcon;
    private ContextMenuStrip TrayIconContextMenu;
    private ToolStripMenuItem CloseMenuItem;

    private void TrayIcon_DoubleClick(object sender, EventArgs e)
    {  
        // How to show Form1 hire.
    }
    //... 
}



 public partial class Form1 : Form
    {
            public Form1()
            {
                InitializeComponent();
            }
           //...
    }

If i instantiate a new class it will give me the new form and all my running processes will be gone example:

private void TrayIcon_DoubleClick(object sender, EventArgs e)
 {  
      Form1 frm = new Form1();
      frm.Show(); //How to show my form?
 }

Not sure if prototype pattern will work here and is there an easier solution?

This is my class:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Drawing;
using System.Windows.Forms;
using System.Windows;

namespace DJBackupDB
{
    class MyApplicationContext : ApplicationContext
    {
        //Component declarations
        private NotifyIcon TrayIcon;
        private ContextMenuStrip TrayIconContextMenu;
        private ToolStripMenuItem CloseMenuItem;
        public static Form1 form;

        public MyApplicationContext()
        {
            Application.ApplicationExit += new EventHandler(this.OnApplicationExit);
            InitializeComponent();
            TrayIcon.Visible = true;
        }

        private void InitializeComponent()
        {
            TrayIcon = new NotifyIcon();

            TrayIcon.BalloonTipIcon = ToolTipIcon.Info;
            TrayIcon.BalloonTipText =
              "I noticed that you double-clicked me! What can I do for you?";
            TrayIcon.BalloonTipTitle = "You called Master?";
            TrayIcon.Text = "My fabulous tray icon demo application";


            //The icon is added to the project resources.
            //Here I assume that the name of the file is 'TrayIcon.ico'
            TrayIcon.Icon = Properties.Resources.logo;

            //Optional - handle doubleclicks on the icon:
            TrayIcon.DoubleClick += TrayIcon_DoubleClick;

            //Optional - Add a context menu to the TrayIcon:
            TrayIconContextMenu = new ContextMenuStrip();
            CloseMenuItem = new ToolStripMenuItem();
            TrayIconContextMenu.SuspendLayout();

            // 
            // TrayIconContextMenu
            // 
            this.TrayIconContextMenu.Items.AddRange(new ToolStripItem[] {
            this.CloseMenuItem});
            this.TrayIconContextMenu.Name = "TrayIconContextMenu";
            this.TrayIconContextMenu.Size = new Size(153, 70);
            // 
            // CloseMenuItem
            // 
            this.CloseMenuItem.Name = "CloseMenuItem";
            this.CloseMenuItem.Size = new Size(152, 22);
            this.CloseMenuItem.Text = "Close the tray icon program";
            this.CloseMenuItem.Click += new EventHandler(this.CloseMenuItem_Click);

            TrayIconContextMenu.ResumeLayout(false);
            TrayIcon.ContextMenuStrip = TrayIconContextMenu;
        }

        private void OnApplicationExit(object sender, EventArgs e)
        {
            //Cleanup so that the icon will be removed when the application is closed
            TrayIcon.Visible = false;
        }

        private void TrayIcon_DoubleClick(object sender, EventArgs e)
        {
            form.Show();
        }

        private void CloseMenuItem_Click(object sender, EventArgs e)
        {
            if (MessageBox.Show("Do you really want to close me?",
                    "Are you sure?", MessageBoxButtons.YesNo, MessageBoxIcon.Exclamation,
                    MessageBoxDefaultButton.Button2) == DialogResult.Yes)
            {
                Application.Exit();
            }
        }
    }
}

Answer 1

You'll have to store your form object somewhere you can access it in the event. Make a class member field for the form just like you have for the NotifyIcon and other things.

private Form1 form;

Then when you create your ApplicationContext, set the form field to the Form1 object.

Then inside the click event handler, you can just call:

this.form.Show();