8

I have an application that opens multiple children widgets as separate windows, something like this: 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?

Virgiliu
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3 Answers3

12

Here is what I have used in __init__ function: QtGui.QShortcut(QtGui.QKeySequence("Ctrl+Q"), self, self.close)

It works smoothly!

xuansamdinh
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3

Try setting the ShortcutContext.

self.actionExit.setShortcutContext(QtCore.Qt.ApplicationShortcut)
Chris B.
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1

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.

James
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