NotifyIcon.Visible Property

Definition

Gets or sets a value indicating whether the icon is visible in the notification area of the taskbar.

public bool Visible { get; set; }

Property Value

true if the icon is visible in the notification area; otherwise, false. The default value is false.

Examples

The following code example demonstrates using the NotifyIcon class to display an icon for an application in the notification area. The example demonstrates setting the Icon, ContextMenu, Text, and Visible properties and handling the DoubleClick event. A ContextMenu with an Exit item on it is assigned to the NotifyIcon.ContextMenu property, which allows the user to close the application. When the DoubleClick event occurs, the application form is activated by calling the Form.Activate method.

using System;
using System.Drawing;
using System.Windows.Forms;

public class Form1 : System.Windows.Forms.Form
{
    private System.Windows.Forms.NotifyIcon notifyIcon1;
    private System.Windows.Forms.ContextMenu contextMenu1;
    private System.Windows.Forms.MenuItem menuItem1;
    private System.ComponentModel.IContainer components;

    [STAThread]
    static void Main() 
    {
        Application.Run(new Form1());
    }

    public Form1()
    {
        this.components = new System.ComponentModel.Container();
        this.contextMenu1 = new System.Windows.Forms.ContextMenu();
        this.menuItem1 = new System.Windows.Forms.MenuItem();

        // Initialize contextMenu1
        this.contextMenu1.MenuItems.AddRange(
                    new System.Windows.Forms.MenuItem[] {this.menuItem1});

        // Initialize menuItem1
        this.menuItem1.Index = 0;
        this.menuItem1.Text = "E&xit";
        this.menuItem1.Click += new System.EventHandler(this.menuItem1_Click);

        // Set up how the form should be displayed.
        this.ClientSize = new System.Drawing.Size(292, 266);
        this.Text = "Notify Icon Example";

        // Create the NotifyIcon.
        this.notifyIcon1 = new System.Windows.Forms.NotifyIcon(this.components);

        // The Icon property sets the icon that will appear
        // in the systray for this application.
        notifyIcon1.Icon = new Icon("appicon.ico");

        // The ContextMenu property sets the menu that will
        // appear when the systray icon is right clicked.
        notifyIcon1.ContextMenu = this.contextMenu1;

        // The Text property sets the text that will be displayed,
        // in a tooltip, when the mouse hovers over the systray icon.
        notifyIcon1.Text = "Form1 (NotifyIcon example)";
        notifyIcon1.Visible = true;

        // Handle the DoubleClick event to activate the form.
        notifyIcon1.DoubleClick += new System.EventHandler(this.notifyIcon1_DoubleClick);
    }

    protected override void Dispose( bool disposing )
    {
        // Clean up any components being used.
        if( disposing )
            if (components != null)
                components.Dispose();            

        base.Dispose( disposing );
    }

    private void notifyIcon1_DoubleClick(object Sender, EventArgs e) 
    {
        // Show the form when the user double clicks on the notify icon.

        // Set the WindowState to normal if the form is minimized.
        if (this.WindowState == FormWindowState.Minimized)
            this.WindowState = FormWindowState.Normal;

        // Activate the form.
        this.Activate();
    }

    private void menuItem1_Click(object Sender, EventArgs e) {
        // Close the form, which closes the application.
        this.Close();
    }
}

Remarks

Since the default value is false, in order for the icon to show up in the notification area, you must set the Visible property to true.

Applies to

Product Versions
.NET Framework 1.1, 2.0, 3.0, 3.5, 4.0, 4.5, 4.5.1, 4.5.2, 4.6, 4.6.1, 4.6.2, 4.7, 4.7.1, 4.7.2, 4.8, 4.8.1
Windows Desktop 3.0, 3.1, 5, 6, 7, 8, 9