Finding whether a widget is popped up or not????
First of all i am not clear of framing proper question.Let me try to post it...
I am developing an chat application on Linux with udp connection...One to one has been successfully completed..i am trying to make one to many i.e,an user can chat with multiple users of app i.e, similar to an gtalk app..when one wants to send data to user, an separate pop up has to come and then chat has to be continued...At the same time if user wants to communicate with other person,an separate window has to be popped up and chat has to be continued...till here i have done successfully...the problem is ,if user responds to my messages ,the received messages has to be put back in respective window only...Here comes the problem...when messages received from him ,i have to check whether dialog is popped up or not..if already popped up ,i should not create a dialog again r else i have to create...How do i do these?????
Help would be highly appreciated...
Waiting for sum interesting answers....:-)
Hii...use it to find out whether your window is pop up/active or not
isActiveWindow : const bool
This property holds whether this widget's window is the active window.
The active window is the window that contains the widget that has keyboard focus (The window may still have focus if it has no widgets or none of its widgets accepts keyboard focus).
When popup windows are visible, this property is true for both the active window and for the popup.
By default, this property is false.
bool isActiveWindow () const
Hope it would help