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PyQt Connect to KeyPressEvent

Certain widgets will allow me to do:

self.widget.clicked.connect(on_click)

but doing:

self.widget.keyPressEvent.connect(on_key)

will fail saying that the object has no attribute 'connect'.

I know that sub-classing the widget and re-implementing the keyPressEvent method will allow me to respond to the event. But how can I .connect() to the keyboard event thereafter, or otherwise said, from a user context?

like image 660
jpcgt Avatar asked Dec 15 '14 00:12

jpcgt


2 Answers

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).

like image 123
ekhumoro Avatar answered Oct 24 '22 08:10

ekhumoro


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'))
like image 5
chris Avatar answered Oct 24 '22 08:10

chris