CODEWITHSUNDEEP
c#.netwinapi
The Mask
The Mask

Reputation: 17427

How to show/hide an application with Visible and ShowInTaskBar as false

How to show/hide an application that's running like: 

Visible = false; 
ShowInTaskBar = false;

using C#?

I tried unsuccessfully, using: 

ShowWindow(handle, SW_SHOWNORMAL);

but it does not show it if the application is running in this above situation.

UPDATE; My scenery:

I have an application(written by me) that when WindowState is FormWindowState.Minimized I hide application of TaskBar and put it in "tray icon mode". I'm using the following method for ensure application single instance:

[STAThread]
static void Main()
{

    bool createdNew;
    Mutex m = new Mutex(true, "...", out createdNew);

    if (!createdNew)
    {
        Process currentProc = Process.GetCurrentProcess();

        foreach (Process proc in Process.GetProcessesByName(currentProc.ProcessName))
        {
            if (proc.Id != currentProc.Id)
            {
                IntPtr handle = currentProc.Handle;
                SetForegroundWindow(handle);
                break;
            }
        }

    }
    else
    {
        Application.SetCompatibleTextRenderingDefault(false);
        Application.EnableVisualStyles();           
        Application.Run(new Form1());
    }
}

the problem is,it works fine for ensure the single instance,but I want show application(exit of tray icon mode) if application is running. I thought to do communication in applications,something like send message from app1(1) to app2, app2 read the message that's 1 and do some action. but I have no idea how do this, ShowWindow() seemed at ago hours the best way to do this,but @Hans Passant pointed some points,that it's not possible. I hope this is clear. Different ways to solve this is very appreciated. Thanks again!

Upvotes: 0

Views: 2247

Answers (2)

PupyMK.1
PupyMK.1

Reputation: 17

I was looking for how to do this without single instance, and I just found the solution.

    [STAThread]
    static void Main()
    {
        string proc = Process.GetCurrentProcess().ProcessName;
        Process[] processes = Process.GetProcessesByName(proc);
        
        // No form has been created
        if (processes.Length <= 1)
        {
            Form1 form = new Form1();
            Application.Run(form);
        }
        // Form has been created
        else
        {
            for (int i = 0; i < processes.Length; i++)
            {
                IntPtr formhwnd = FindWindow(null, "Form1");// get HWND by the text name of form

                // Use WIN32 methods
                int style = GetWindowLong(formhwnd, GWL_EXSTYLE);// Get EX-Style of the window
                style |= WS_EX_APPWINDOW;// Add the APP style that shows icon in taskbar
                style &= ~(WS_EX_TOOLWINDOW);// Delete the tool style that does not show icon in taskbar

                SetWindowLong(formhwnd, GWL_EXSTYLE, style);// Set the EX-Style
                ShowWindow(formhwnd, SW_SHOW);// Show the Window
                SwitchToThisWindow(formhwnd, true);// Focus on the window
            }
        }
    }

If you want to show/hide a window from another app. This can still be a reference.Just get the handle of that window and use the win32 methods (import from C++ dll) to set the window styles.

Upvotes: 0

Hans Passant
Hans Passant

Reputation: 941635

Changing the ShowInTaskbar property changes the Handle value. It is one of several Form class properties that can only be specified in the native CreateWindowEx() call and can't be changed later. So changing the property requires Winforms to re-create the window. And that gives it a different handle, making it very likely that the ShowWindow() call uses the wrong value.

You didn't find out that this was the problem because you are not checking the ShowWindow() return value. Very important when you pinvoke Windows calls, you don't have a friendly .NET exception to whack you over the head when the call failed.

Upvotes: 3

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