Open only one instance of the dialog

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VS2005

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I have one issue regarding to the dialog box. I have one menu, after clicked on this a dialog is popped up. When I clicked again on the same menu without closing the last opened dialog an another instance of same dialog is opened. This happens till I close all the created instance of the dialog. How I found that the dialog's instance is already run or not or how to detect that the dialog is opened or not? Please somebody help me

Posted 11-Oct-12 20:44pm

Vaibhav_J_Jaiswal

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[no name] 12-Oct-12 7:47am

Simple, show the dialog modal then you will not have this problem.

2 solutions

Solution 1

Use FindWindow() to get handle to your window.

C++

HWND hDlgExists = ::FindWindow(0,"MyDialogTitle"); // hDlgExists will be NULL if dlg is not exist.

http://msdn.microsoft.com/en-us/library/windows/desktop/ms633499%28v=vs.85%29.aspx

Posted 11-Oct-12 21:00pm

Santhosh G_

Comments

Vaibhav_J_Jaiswal 12-Oct-12 3:09am

I used this but in future if the name of the dialog is changed then this method is not useful. Is there any another way in which I pass the handle of that dialog and then it detect that this dialog is opened or not?

Santhosh G_ 12-Oct-12 3:32am

// The handle to be tested.
HWND hWindow = GetSafeHwnd();

// Check whether the window is still there.
BOOL bWindowAlive = IsWindow( hWindow );

Vaibhav_J_Jaiswal 12-Oct-12 4:05am

I got the handle of the dialog as null. How I get the handle of the window? I used the below code:

void CVTXMIOCaptureDlg::OnContextViewsetup()
{
CEngineSetupDialog *dlg = new CEngineSetupDialog();
HWND hWindow = dlg->GetSafeHwnd();
BOOL bWndAlive = ::IsWindow(hWindow);
if(!bWndAlive)
{
dlg->DoModal();
}
}

Santhosh G_ 12-Oct-12 6:14am

Simply creating a CDialog instance will not give a valid window handle.
TestDlg* pDlg = new TestDlg();
HWND hNullHandle = pDlg->GetSafeHwnd(); // Handle will be NULL
pDlg->Create( IDD_DIALOG_ID, 0 );
HWND hValidaHandle = pDlg->GetSafeHwnd(); // Handle will be Valid
pDlg->ShowWindow( SW_SHOW );

Santhosh G_ 12-Oct-12 5:36am

If it is DoModal() to display the dialog, how you will get more than one dialog ?
Are you creating new process, or new thread to display new dialog ?

Vaibhav_J_Jaiswal 12-Oct-12 7:15am

No, I can't use or create any process or thread. It is just simple dialog whenever user select the menu it shows that dialog. It is DoModal(), but whenever I click on the menu it creates the new one dialog which is wrong. This menu item is handled from a tray icon, that's why user select that menu item again and again and the new dialog instance get created. How to resolved it please tell me.

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Santhosh G_ · Accepted Answer · 2012-10-12T01:40:00

Solution 2

C++

bool bAlreadyRunning;

HANDLE hMutexOneInstance = ::CreateMutex( NULL, FALSE, _T("Your_Dialog_Name"));

bAlreadyRunning = ( ::GetLastError() == ERROR_ALREADY_EXISTS ||
                    ::GetLastError() == ERROR_ACCESS_DENIED);

if( !bAlreadyRunning )
{
  // Initialize Dialog and Call DoModal()
}

Posted 12-Oct-12 1:40am

Santhosh G_

Comments

Vaibhav_J_Jaiswal 12-Oct-12 8:12am

instead of using mutex I directly used FindWindow() method because both are used the Name of the dialog instead of Handle.

Santhosh G_ 12-Oct-12 8:22am

Its not required to have the same name of Dialog name.
Any unique name to indicate your dialog. It should be unique. If anyone else opened a mutex with same name, your dialog will not appear.

Vaibhav_J_Jaiswal 12-Oct-12 8:33am

I used the above code it creates an issue: As first time it creates the window then if I again select the menu it didn't allow to create dialog again. But if I selects the Apply or Cancel button, the dialog closed and after that I select the menu then it didn't allow to create the dialog. It happens till I exit the tray icon. When I start the application again it work and did the same thing as I above explained.

Santhosh G_ 12-Oct-12 8:48am

After DoModal(), please call ReleaseMutex() and CloseHandle() to close the mutex instance.

Vaibhav_J_Jaiswal 12-Oct-12 9:15am

Thanx for your help