How to make a form visible by click in its windows tray icon

Question

In somewhere of this page I read that the best way to "override" the minimize method is to use onResize() event. I've done it and it works! I've coded that when the form is minimized it turns no visible and put an icon in the windows tray. As yet everything is working fine, but I've also programmed (or tried it at least) that when the icon in windows tray is clicked, the form turns visible again, but it doesn't work and I don't know why.

I've tried to code both events (Click and MouseClick), but the code is the same in both and it still doesn't work, so I must be doing something wrong, but obviusly I don't know what.

Remember that what I want is to code the click event on windows tray icon, not on task bar icon, so maybe that's why it isn't working, maybe is another event.

Here is my code:

private void onResize(object sender, EventArgs e)
    {
        this.ShowInTaskbar = false;
        notifyIcon1.Visible = true;
        this.Visible = false;
    }

    private void notifyIcon_MouseClick(object sender, MouseEventArgs e)
    {
        this.ShowInTaskbar = true;
        this.Visible = true;
    }

    private void notifyIcon_Click(object sender, EventArgs e)
    {
        this.ShowInTaskbar = true;
        this.Visible = true;
    }

Thanks.

EDIT: the problem was that I just assigned the image to the icon on the Form constructor like notifyIcon1.Icon = new Icon("greenCircle.ico"); but I didn't initialize any image on the notifyIcon properties... By initializing the image on notifyIcon properties it worked fine!

score 1 · Accepted Answer · answered Nov 01 '14 at 19:30

1

You probably don't want to hide it when they resize it, just when they minimize the Form:

private void onResize(object sender, EventArgs e)
{
    if (WindowState == FormWindowState.Minimized)  // only hide if minimizing the form
    {
        this.ShowInTaskbar = false;
        notifyIcon1.Visible = true;
        this.Visible = false;
    }
}

To show the minimized Form again, you'll need to restore it:

private void notifyIcon_MouseClick(object sender, MouseEventArgs e)
{
    this.ShowInTaskbar = true;
    this.Visible = true;

    WindowState = FormWindowState.Normal;
}

answered Nov 01 '14 at 19:30

Grant Winney

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Right, it never comes back again. I did it as you said "FormWindowState.Normal". However, just in case, I put "this.WindowState", but nothing changes. – Drumnbass Nov 01 '14 at 19:54
Damn, how didn't I think about do that... you definitely nailed it. The event isn't actually firing. Now the point is to know why. As I said, could it be another event? – Drumnbass Nov 01 '14 at 20:09
Nope, I actually look over it many times, and I've done it again, just in case, but it's correctly suscribed. `this.notifyIcon1.MouseClick += new System.Windows.Forms.MouseEventHandler(this.notifyIcon_MouseClick);` – Drumnbass Nov 01 '14 at 20:29
Right, that is from the `Designer.cs` file. By the way, I have found this another thread that could be the solution, but I don't understand it at all, it is beyond my knowledge. What do you think? http://stackoverflow.com/questions/2595474/winforms-taskbar-icon-click-event-not-firing?rq=1 – Drumnbass Nov 01 '14 at 20:39
Waoh. I finally found the problem... I just assigned the image to the icon on the Form constructor like `notifyIcon1.Icon = new Icon("greenCircle.ico");` but I didn't initialize any image on the notifyIcon properties... I really don't know what does it have to do with that, but the point is that it worked! Anyway, your help completed my code, so thanks! ;) – Drumnbass Nov 01 '14 at 20:59

score 1 · Answer 2 · answered Nov 01 '14 at 21:22

    this.ShowInTaskbar = true;
    this.Visible = true;

This doesn't do what you hope it does. A tricky problem in Windows that has bedeviled many programmers. The ShowInTaskbar property is "special", very unlikely most other Form properties. It is specified in the underlying CreateWindowEx() winapi function as a style bit. There are other properties like that, like FormBorderStyle, ControlBox, LeftToRight. Each corresponding with a style bit.

Trouble is, changing the property requires Winforms to perform considerable gymnastics, it has the destroy the native window and re-create it again. That tends to have side-effects, you found one.

Just swap the two statements and you'll see it working very differently, now it does manage to make the window visible again. But still not smoothly, it doesn't always manage to get into the foreground.

You might be puzzled at what looks like a massive bug. The real problem is that you are doing something that most other programmers never do, always a good way to get a problem that nobody else has. You don't actually have to change the ShowInTaskbar property at all. The taskbar button is already automatically invisible when you hide your window, no need to help at all.

Delete all ShowInTaskbar assignments in your program to fix your problem.

How to make a form visible by click in its windows tray icon

2 Answers2