[Qt-interest] QDockWidget toggle visibility styled

[Christian Lackas]

Hi Everybody,

I have a bunch of tabbed QDockWidgets, and corresponding buttons in a
QToolBar that should toggle their visibility. If a QDockWidgets is not
visible, I show() and raise() it (the latter one to select its tab),
which works find.

My problem starts, when a QDockWidgets is already visible, but a
different tab is shown. In this case isVisible() is still true, but
rather than hide() I would like to raise() it to  The Widget should only
be hidden if it is actually visible to the user.

I already tried

    connect(toolBarAction, SIGNAL(triggered()),
        dockWidget->toggleViewAction(), SLOT(trigger()));

(which does not help and also does not contain the raise() I need) as
well as

    connect(toolBarAction, SIGNAL(triggered()), SLOT(toggleDockView()));

    void MainWin::toggleDockView() {
        if (dockWidget->isVisible()) {
            if (*is actually visible, e.g. tab shown*) {    // !!!
                dockWidget->hide();
            } else {
                dockWidget->raise();
            }
        } else {
            dockWidget->show(); dockWidget->raise();
        }
    }

If I use (in above isVisible branch):

    const int cnt = dockWidget->parentWidget()->children().size();
    if (cnt < 2) return dockWidget->hide();
    const int idx =
        dockWidget->parentWidget()->children().indexOf(dockWidget);
    if (idx == cnt -1)
        dockWidget->hide();
    else
        dockWidget->raise();

it works for the case described above (widget is topmost), but not for
the case the the QDockWidget is the last one visible (in this case, the
idx is != 0, but also not the last one), which means I have to click the
toggle button twice to hide the last dock widget.

Do I have to add just another exception for this case, or is there a
simpler way to achieve this?

Any insight is appreciated.

Thanks,
 Christian