PyQt Connect to KeyPressEvent

Question

Certain widgets will allow me to do:

self.widget.clicked.connect(on_click)

but doing:

self.widget.keyPressEvent.connect(on_         


        

Answer 1

The way I've done this in the past is (it is a work around), where this is in the sender and the receiver declares/connects to the signal.

def keyPressEvent(self, event):
    if type(event) == QtGui.QKeyEvent:
        if event.key() == QtCore.Qt.Key_Space:
            self.emit(QtCore.SIGNAL('MYSIGNAL'))

Answer 2

Create a custom signal, and emit it from your reimplemented event handler:

class MyWidget(QtGui.QWidget):
    keyPressed = QtCore.pyqtSignal(int)

    def keyPressEvent(self, event):
        super(MyWidget, self).keyPressEvent(event)
        self.keyPressed.emit(event.key())
...

def on_key(key):
    # test for a specific key
    if key == QtCore.Qt.Key_Return:
        print('return key pressed')
    else:
        print('key pressed: %i' % key)

self.widget.keyPressed.connect(on_key)

(NB: calling the base-class implementation is required in order to keep the existing handling of events).