Two (probably) very simple questions

Sorren

Thanks, everything works now.

Stumbino

Has anyone gotten a button to click for the pyqt gui? This is a GPIO Output signaling button is clicked. Does anyone have a solution to this?

jazzycamel

@Stumbino
Not sure that I understand your question: do you mean you want a Qt signal to be emitted when a a GPIO input is changed/toggled?

Stumbino

@jazzycamel Most of the resources online are only discussing pyqt and gpio output(ex: turning a lightbulb on). Has anyone gotten pyqt to work as a GPIO button signalling the computer that the button is pressed as a GPIO input (sorry messed that up previously)? The only information that I have found so far is making a thread (qthread) so looping can be processed separately, but I have not seen an example of this working for GPIO input . An example would be greatly appreciated! Thanks!

jazzycamel

@Stumbino

Something like the following should do what you want (if I'm understanding you correctly):

import RPi.GPIO as GPIO
GPIO.setmode(GPIO.BCM)

from functools import partial
from PyQt5.QtCore import QObject, pyqtSignal
pyqtWrapperType=type(QObject)

class Singleton(pyqtWrapperType):
    _instances={}
    def __call__(cls, *args, **kwargs):
        if cls not in cls._instances:
            cls._instances[cls]=super(Singleton, cls).__call__(*args, **kwargs)
        return cls._instances[cls]

class QGPIO(QObject, metaclass=Singleton):
    interrupted=pyqtSignal(int,int)

    @staticmethod
    def _callback(pin, event):
        QGPIO().interrupted.emit(pin,event)

    def register(self, pin, event, bounceTime=300):
        callback=partial(self._callback, pin, event)
        GPIO.add_event_detect(pin, event, callback=callback, bouncetime=bounceTime)

if __name__=="__main__":
    from sys import argv, exit
    from PyQt5.QtCore import pyqtSlot
    from PyQt5.QtWidgets import QApplication, QWidget

    class Widget(QWidget):
        def __init__(self, parent=None, **kwargs):
            super().__init__(parent, **kwargs)

            GPIO.setup(17, GPIO.IN, pull_up_down=GPIO.PUD_UP)

            QGPIO().interrupted.connect(self.interrupted)
            QGPIO().register(17, GPIO.FALLING)

        @pyqtSlot(int,int)
        def interrupted(self, pin, event):
            print("Interrupted:", pin, event)

    a=QApplication(argv)
    w=Widget()
    w.show()
    exit(a.exec_())

Basically(!), I create a singleton QObject() that has a callback that can be registered with the RPi.GPIO interrupt system such that, when an edge change is detected, the QGPIO() class will trap the interrupt via _callback and then emit a signal interrupted(int,int) which will inform a slot of which pin has changed and by which event type.

Just a warning/disclaimer: I've tested all the PyQt bits, but I had to stub the RPi.GPIO bits as I don't have a Pi handy right this second.

Stumbino

@jazzycamel What do I change to make this work for pyqt4? I am currently using a RaspberryPi 0 and it looks like pyqt5 doesnt have full support for arm v6 cpu.

jazzycamel

@Stumbino
Not much I shouldn't have thought...

Change the PyQt5 imports to PyQt4 and use the Python 2 (I'm assuming you are using Python 2 with PyQt4?) metaclass syntax i.e.:

class QGPIO(QObject):
    __metaclass__=Singleton

Should work fine. I don't have a PyQt4 setup to test I am afraid, but if you have any issues just post 'em here :)

Stumbino

@jazzycamel I got all the libraries switched over to pyqt4, but I get these two errors when I run the program.

jazzycamel

@Stumbino Replace the line containing super. __init__ with:

QWidget.__init__(self, parent, **kwargs)

Stumbino

@jazzycamel That worked! How would I load a UI file in the Widget class? I tried using uic.loadUI(), but that seems to override the previous gui.