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_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?

Correct me if I'm wrong, but keyPressEvent is not a slot so it can't be connected. You have to handle it using any of the event handlers. — f.rodrigues, Dec 15 '14 at 01:45

ekhumoro · Accepted Answer · 2018-05-01T15:56:03.057

18

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).

edited May 01 '18 at 15:56

answered Dec 15 '14 at 03:43

ekhumoro

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1

Do you not need to overide `__init__` and call the super's `__init__`? – wesanyer Jan 25 '16 at 20:23
1

@wesanyer. No. Python always automatically calls `__init__` on each base-class, regardless of whether it has been overidden or not. But of course, if you **do** override `__init__`, you will then have to explicitly call the base-class implementation as well (because it will no longer be called automatically). – ekhumoro Jan 25 '16 at 23:12
1

what is on_key in this context? – Szabolcs Dombi Jul 11 '16 at 09:03
@DombiSzabolcs. It's a slot that is called when the signal is emitted. In PyQt/PySide, a slot can be any python callable. – ekhumoro Jul 11 '16 at 14:11

score 5 · Answer 2 · answered Dec 15 '14 at 00:59

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'))

PyQt Connect to KeyPressEvent

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