[FIXED] Setting up QStackedWidget properly, signal cannot see function/slot

Issue

I am working on developing a GUI with PyQt5. This is my first step into OOP, and I'm trying to teach myself as I go. I'm struggling with understanding when classes inherit methods/attributes etc and what methods they have available -- I guess it is a scope-related question? I have produced a MWE to my GUI below. In total, there will be many more pages and signals/slots.

What I want:

The stack should initialize with the "MainMenu" widget/object showing (left image below). Clicking on "Next Page" button should switch the stack order to put the "OtherPage" widget/object on top (right image below). I am creating each page as a class, thinking this would be a good way to organize my project. Is this good or bad practice?

What happens now:

The GUI works (initializes) if the line nextPg.clicked.connect(self.drawOtherPage()) is commented out, but of course then clicking on the button does nothing. I can switch the initial stack order so that "other" widget is on top of the stack and it shows up fine, so I think that class is also working. When the above line is included in the code, the following error is thrown:

in __init__
     nextPg.clicked.connect(self.drawOtherPage())
AttributeError: 'MainMenu' object has no attribute 'drawOtherPage'

What I've tried

I thought that the call to super() was supposed to allow the child class (in this case MainMenu) to inherit the methods from the parent class (RootInit). Therefore, I would think this should make the drawOtherPage method available to the button connect signal. Obviously, the error isa result of the method not being available.

What am I doing wrong? Should I be creating these "page" widgets in methods instead? Do they need to be under the RootInit class or can they live in the top level of the .py file? I'm trying to follow best practices as the project will become pretty large in the end. Fortunately, most of it should be pages with variations based on what buttons were clicked to get there -- I therefore thought classes would be helpful. Please be harsh on the code and my python/PyQt vernacular, trying to learn -- thanks!

import sys, os
from PyQt5.QtWidgets import *
from PyQt5 import QtGui, QtCore

class RootInit(QMainWindow):
    # root window
    def __init__(self, parent=None):
        QMainWindow.__init__(self)
        self.root = QWidget()
        self.stack = QStackedWidget()
        rootLayout = QVBoxLayout()
        rootLayout.addWidget(self.stack)
        self.root.setLayout(rootLayout)
        self.setCentralWidget(self.root)

        self.initializeGUI()

    def initializeGUI(self):
        self.main = MainMenu(self) # build MainMenu (class)
        self.other = OtherPage(self) # build OtherPage (class)
        self.stack.addWidget(self.main)
        self.stack.addWidget(self.other)

    def drawMain(self):
        self.stack.setCurrentIndex(0)

    def drawOtherPage(self):
        self.stack.setCurrentIndex(1)


class MainMenu(QWidget):
    # class for main menu
    def __init__(self, parent=None):
        QWidget.__init__(self, parent)
        super().__init__()

        mainLayout = QGridLayout() # layout for entire main menu
        quitBtn = QPushButton("Quit")
        quitBtn.clicked.connect(QtCore.QCoreApplication.instance().quit)
        nextPg = QPushButton("Next page")
        nextPg.clicked.connect(self.drawOtherPage())

        mainLayout.addWidget(quitBtn, 0, 0)
        mainLayout.addWidget(nextPg, 0, 1)
        self.setLayout(mainLayout)


class OtherPage(QWidget):
    # class for another menu
    def __init__(self, parent=None):
        QWidget.__init__(self, parent)
        label = QLabel("test label")
        layout = QGridLayout() #
        layout.addWidget(label, 0, 0)
        self.setLayout(layout)


if __name__ == '__main__':
    app = QApplication(sys.argv)
    root = RootInit()
    root.setWindowTitle("Title")
    root.show()
    sys.exit(app.exec_())

Solution

Your code has the following errors:

• The variable self refers to the same instance of the class, in your case self refers to an instance of MainMenu, and if we observe MainMenu it does not have any drawOtherPage() method.

• Another mistake in your case is to call the parent's constructor twice:

---

class MainMenu(QWidget):
    # class for main menu
    def __init__(self, parent=None):
        QWidget.__init__(self, parent)
        super().__init__()

In the first constructor you are assigning a parent, and in the second, you are not. To clarify in python there are several ways to call the parent's constructor:

    QWidget.__init__(self, parent)
    super(MainMenu, self).__init__(parent)
    super().__init__(parent)

so you should only use one of them.

• Another error is that a signal is connected through the name of a function, the function must not be evaluated using parentheses

• and for the last use of functions or methods that involve several objects should be done in a place where both objects can access, in your case you can take advantage of what you are going to RootInit as parent of MainMenu: self.main = MainMenu(self), and access the connection to that element through the method parent().

All of the above entails modifying the MainMenu class to the following:

class MainMenu(QWidget):
    def __init__(self, parent=None):
        QWidget.__init__(self, parent)

        mainLayout = QGridLayout() # layout for entire main menu
        quitBtn = QPushButton("Quit")
        quitBtn.clicked.connect(QtCore.QCoreApplication.instance().quit)
        nextPg = QPushButton("Next page")
        nextPg.clicked.connect(self.parent().drawOtherPage)

        mainLayout.addWidget(quitBtn, 0, 0)
        mainLayout.addWidget(nextPg, 0, 1)
        self.setLayout(mainLayout)

Answered By - eyllanesc