How to show the menu in QPushButton without blocki

问题:

I use Qt4 QPushButton with QMenu in it (set by setMenu()). I need to show this menu when some unrelated event occurs.

Method QPushButton::showMenu() does this, but it blocks until user closes the menu. QMenu::show() also does this, but it shows the menu in the top left corner of the screen.

How can I programmatically make the menu show up properly positioned, and without blocking?

回答1:

QMenu is QWidget. So you can call move() before show().

回答2:

You can use

void QMenu::popup ( const QPoint & p, QAction * atAction = 0 )

So

Displays the menu so that the action atAction will be at the specified global position p. To translate a widget's local coordinates into global coordinates, use QWidget::mapToGlobal().

回答3:

No, I didn't like the suggested solutions, because QPushButton is supposed to manage the menu position, not the caller.

So I decided to post mouse down/up events to this QPushButton widget, simulating what the user does. This did the trick. This is a hack to compensate for the missing functionality in Qt.

void simulateMouseClick(QWidget *widget) {
  QPoint pos(widget->width()/2, widget->height()/2);
  QMouseEvent *evtDown = new QMouseEvent(QEvent::MouseButtonPress, pos, Qt::LeftButton, Qt::LeftButton, Qt::NoModifier);
  QMouseEvent *evtUp   = new QMouseEvent(QEvent::MouseButtonRelease, pos, Qt::LeftButton, Qt::LeftButton, Qt::NoModifier);
  (void) QApplication::postEvent(widget, evtDown);
  (void) QApplication::postEvent(widget, evtUp);
}