I have an application that opens multiple children widgets as separate windows, something like th开发者_开发问答is: window1 opens window 2 which opens window 3 (simplified form).
In the main window I have set CTRL+Q as the quit shortcut. Below is a stripped down example of the main class.
class MainWindow(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
self.actionExit = QtGui.QAction(_('E&xit'),self)
self.actionExit.setShortcut('Ctrl+Q')
self.actionExit.setStatusTip(_('Close application'))
self.connect(self.actionExit, QtCore.SIGNAL('triggered()'), QtCore.SLOT('close()'))
Right now if I open the third child and push CTRL+Q nothing happens. Is there a way so that the children inherit the shortcut key for quit or to make the shortcut global or do I have to declare it in each of them?
Here is what I have used in __init__
function:
QtGui.QShortcut(QtGui.QKeySequence("Ctrl+Q"), self, self.close)
It works smoothly!
Try setting the ShortcutContext.
self.actionExit.setShortcutContext(QtCore.Qt.ApplicationShortcut)
You can also just set a shortcut for your QAction directly:
self.actionExit.setShortcut(QtGui.QKeySequence("Ctrl+Q"))
The only difference between this example and your code is that the Ctrl+Q
is first cast to QtGui.QKeySequence
.
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