CODEWITHSUNDEEP
pythonpython-3.xpyqtpyqt5
Joubert Lucas
Joubert Lucas

Reputation: 163

Python3 PyQt5 setEnabled of a QAction causing crashes

For a project I'm creating a GUI using Python 3 and PyQt5. Because it has to be usable by people outside of my immediate team, I want to disable actions on the menu until they've already filled out some forms in other parts of the program (e.g. disabling the final solution view when they haven't set up the initial data connection). The issue is that when I try to call the QAction's setEnabled function outside of the function that created it (but still inside the overall class), it's causing my script to crash with no error code, so I'm having trouble understanding the issue. In the snipit below, I'm trying to set the "View Solution" menu option as true. There are some more options in that menu, but I deleted them here to make it more easy to read.

The code is structured something like this:

import sys
from PyQt5.QtWidgets import QMainWindow, QAction, qApp, QApplication, QMessageBox, QStackedLayout


class MediaPlanner(QMainWindow):

    def __init__(self):
        super().__init__()

        self.initUI()

    def initUI(self):

        # Menu bar example from: zetcode.com/gui/pyqt5/

        exitAction = QAction('&Exit', self)
        exitAction.setShortcut('Ctrl+Q')
        exitAction.setStatusTip('Exit application')
        exitAction.triggered.connect(self.close)

        newProject = QAction('&New Project', self)
        newProject.setShortcut('Ctrl+N')
        newProject.setStatusTip('Start A New Project')
        newProject.triggered.connect(self.createNewProject)

        openProject = QAction('&Open Project',self)
        openProject.setShortcut('Ctrl+O')
        openProject.setStatusTip('Open A Saved Project')
        openProject.setEnabled(False)

        viewSolution = QAction('&View Solution',self)
        viewSolution.setStatusTip('View the Current Solution (If Built)')
        viewSolution.setEnabled(False)

        self.statusBar()

        menubar = self.menuBar()

        filemenu = menubar.addMenu('&File')
        filemenu.addAction(newProject)
        filemenu.addAction(openProject)
        filemenu.addAction(exitAction)

        viewmenu = menubar.addMenu('&View')
        viewmenu.addAction(viewSolution)

        self.setGeometry(300,300,700,300)
        self.setWindowTitle('Menubar')

        self.show()

    def createNewProject(self):
        print('Project Created')
        self.viewSolution.setEnabled(True)



if __name__ == '__main__':

    app = QApplication(sys.argv)
    gui = MediaPlanner()
    sys.exit(app.exec_())

Upvotes: 1

Views: 2468

Answers (1)

eyllanesc
eyllanesc

Reputation: 244301

The problem is that viewSolution is a variable, but it is not a member of the class so you will not be able to access it through the self instance. One possible solution is to make viewSolution member of the class as shown below:

self.viewSolution = QAction('&View Solution',self)
self.viewSolution.setStatusTip('View the Current Solution (If Built)')
self.viewSolution.setEnabled(False)
...
viewmenu.addAction(self.viewSolution)

Another possible solution is to use the sender() function, this function returns the object that emits the signal, using the following:

def createNewProject(self):
    print('Project Created')
    self.sender().setEnabled(True)

Upvotes: 1

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