Qt-interest Archive, April 2008
determine if a dock widget is docked with another specific widget?
Message 1 in thread
I have a case where I need to pop to the front a dock widget if
something in another dock is clicked, but I ONLY want to do this if the
two docks in question are not also docked together.
The upshot is I need a "isDockedWith()" method on either a QDockWidget
or the QMainWindow, but no such function exists (Qt 4.3.x).
Has anyone come up with a work-around or implementation for such a function?
--
[ signature omitted ]
Message 2 in thread
On torsdag den 3. April 2008, Paul Miller wrote:
> I have a case where I need to pop to the front a dock widget if
> something in another dock is clicked, but I ONLY want to do this if the
> two docks in question are not also docked together.
>
> The upshot is I need a "isDockedWith()" method on either a QDockWidget
> or the QMainWindow, but no such function exists (Qt 4.3.x).
>
> Has anyone come up with a work-around or implementation for such a
> function?
Perhaps parentWidget() on the dock widgets, or search for the dock widgets in
QMainWindow with window->findChildren<QDockWidget*>()?
Bo.
--
[ signature omitted ]
Message 3 in thread
Bo Thorsen wrote:
> On torsdag den 3. April 2008, Paul Miller wrote:
>> I have a case where I need to pop to the front a dock widget if
>> something in another dock is clicked, but I ONLY want to do this if the
>> two docks in question are not also docked together.
>>
>> The upshot is I need a "isDockedWith()" method on either a QDockWidget
>> or the QMainWindow, but no such function exists (Qt 4.3.x).
>>
>> Has anyone come up with a work-around or implementation for such a
>> function?
>
> Perhaps parentWidget() on the dock widgets, or search for the dock widgets in
> QMainWindow with window->findChildren<QDockWidget*>()?
Yeah, I tried that, as well as comparing parent layouts. Guess I'll take
a trip through the Qt source.
--
[ signature omitted ]
