Python: decoupling code from PyQt5 UI Code

Question

I'm doing my first steps with PyQt5. I created a simple dialog "form_main" with only 1 menu: "File|Quit":

import sys
from PyQt5.QtWidgets import QApplication, QMainWindow, QAction
from PyQt5.QtGui import QIcon

class GUI(QMainWindow):
    def __init__(self):
        super().__init__()
        self.initUI()

    def initUI(self):
        self.setWindowTitle('MyApp')
        self.resize(400, 300)
        self.statusBar().showMessage('Welcome')
        self.add_menus()

    def add_menus(self):
        menubar = self.menuBar()
        file_menu = menubar.addMenu('&File')

        exit_action = QAction('&Quit', self)
        file_menu.addAction(exit_action)

Now I want to access this code from another class. In that class I want to define a function that's called when I click on "File|Quit". In a second file I wrote:

from frm_main import GUI
from PyQt5 import QtWidgets
from PyQt5.QtWidgets import QApplication

def exit_action():
    self.close

if __name__ == "__main__":
    import sys
    app = QtWidgets.QApplication(sys.argv)
    MainWindow = QtWidgets.QMainWindow()
    ui = GUI()
    # ui.exit_action.triggered.connect(exit_action)
    ui.show()
    sys.exit(app.exec_())

This code correctly shows my dialog, but when I remove the comment where I try to link the "File|Quit" menu with the function exit_action, it breaks.

Any hint how I can connect the menu to the function?

score 2 · Accepted Answer · answered Apr 15 '18 at 23:13

Your code has the following errors:

only the attributes can be accessed from outside the class, in your case exit_action is not, in order to be an attribute you must precede it self.
in the case of the function exit_action is not defined self, in the classes the first attribute of every method is the instance, which by convention is given the name of self, in the case exit_action should be the widget, for this we can use a lambda function, but you also have to invoke the close function using parentheses.

frm_main.py

import sys
from PyQt5.QtWidgets import QApplication, QMainWindow, QAction
from PyQt5.QtGui import QIcon

class GUI(QMainWindow):
    def __init__(self):
        super().__init__()
        self.initUI()

    def initUI(self):
        self.setWindowTitle('MyApp')
        self.resize(400, 300)
        self.statusBar().showMessage('Welcome')
        self.add_menus()

    def add_menus(self):
        menubar = self.menuBar()
        file_menu = menubar.addMenu('&File')

        self.exit_action = QAction('&Quit', self)
        file_menu.addAction(self.exit_action)

main.py

from frm_main import GUI
from PyQt5 import QtWidgets
from PyQt5.QtWidgets import QApplication

def exit_action(widget):
    widget.close()

if __name__ == "__main__":
    import sys
    app = QtWidgets.QApplication(sys.argv)
    MainWindow = QtWidgets.QMainWindow()
    ui = GUI()
    ui.exit_action.triggered.connect(lambda: exit_action(ui))
    ui.show()
    sys.exit(app.exec_())

Wow, thank you very much. That worked. Seems I still have a LOT to learn. Next thing I'll have to read about will be lambda, I guess. — Anumi, Apr 16 '18 at 05:51

Python: decoupling code from PyQt5 UI Code

1 Answers1