Python在关闭GUI应用程序时意外退出

Question

I am trying to write a simple menu driven GUI program.

Following are the environment:

1. PyQt5 ver.5.10.1
2. Python 3.6.4
3. Mac OS Sierra

PyQt5 was installed using pip3.

Here is the code I am using:

from PyQt5.QtWidgets import (QMainWindow, QApplication,
                         QWidget, QPushButton, QAction)
from PyQt5.QtGui import QIcon
from PyQt5.QtCore import pyqtSlot

class myApp(QMainWindow):

    def __init__(self):
        super().__init__()
        self.title = 'Some App'
        self.left = 10
        self.top = 10
        self.width = 480
        self.height = 260
        self.initUI()

    def initUI(self):
        self.setWindowTitle(self.title)
        self.setGeometry(self.left, self.top, self.width, self.height)
        self.statusBar()
        pkMenuBar = self.menuBar()
        mnuFile = pkMenuBar.addMenu("File")
        mnuFile.addAction("Create New")
        mnuQuit = QAction(QIcon("ico1.png"), " Quit", self)
        mnuQuit.setShortcut("Ctrl+Q")
        mnuQuit.setStatusTip("Quit Application")
        mnuFile.addAction(mnuQuit)

        mnuFile.triggered[QAction].connect(self.triggerAct)

        self.show()

    def triggerAct(self, x):
        if x.text() == "Create New":
            print("Trigger Create New...")
        elif x.text() == " Quit":
            mnuQuit.triggered.connect(self.close)

if __name__ == '__main__':
    import sys
    app = QApplication(sys.argv)
    ex = myApp()
    sys.exit(app.exec_())

On executing the code (using IDLE) the GUI loads and the menu items (Actions) are also working as expected.

When the menu item "Quit" is called the app closes as well as the python icon (from the system tray). But after about 5-10 seconds I keep getting a message saying "Python quit unexpectedly".

I have been trying to figure out the problem using possible solns. by following leads on the net (like sys.exit(), app.quit() ) but every time I face the same result.

I have coding experience on ABAP/4 and VB.Net but this is very early stages for me so far as GUI coding on Python/PyQt is concerned.

Will appreciate if a lead is provided so that I may progress in this new endeavor of mine.

Thanks

Answer 1

The problem is really simple: mnuQuit is a variable not a member of the class so it can not be accessed by other methods of the class, plus I see it unnecessary to use that line of code, just call close() :

def triggerAct(self, x):
    if x.text() == "Create New":
        print("Trigger Create New...")
    elif x.text() == " Quit":
        self.close()
        # mnuQuit.triggered.connect(self.close)

---

It is not appropriate to call exit() within the GUI because the GUI has to free memory and if you call exit() it does not give you the time to do it. with close() the GUI closes correctly.

Answer 2

 def triggerAct(self, x):
    if x.text() == "Create New":
        print("Trigger Create New...")
    elif x.text() == " Quit":
        exit()

I think exit() will work fine in your case.