我有一个Application Context类,在其中我有一些方法将Form1类发送到system-tray,我的问题是如何在系统托盘中运行时显示已经实例化的类From1,例如:
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();
}
//...
}
如果我实例化一个新类,它将为我提供新表单,并且所有正在运行的进程都将消失为例:
private void TrayIcon_DoubleClick(object sender, EventArgs e)
{
Form1 frm = new Form1();
frm.Show(); //How to show my form?
}
不确定原型模式是否可以在这里工作,是否有更简单的解决方案?
这是我的班级:
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();
}
}
}
}
答案 0 :(得分:1)
您必须将表单对象存储在可以在事件中访问它的位置。为表单创建一个类成员字段就像你对NotifyIcon和其他东西一样。
private Form1 form;
然后在创建ApplicationContext时,将表单字段设置为Form1对象。
然后在click事件处理程序中,您只需调用:
this.form.Show();