如何销毁QApplication，然后在不退出python脚本的情况下运行新的QApplication？]

Question

我想创建一个QApplication，然后使用键盘快捷键退出它。然后，python脚本应调用另一个QApplication。

我目前的问题是，当第二个QApplication将要运行时，我收到此错误：

app2 = QApplication()
RuntimeError: Please destroy the QApplication singleton before creating a new QApplication instance.
我具有以下结构：

| main.py
| Q1.py
| Q2.py
这是main.py：

import Q1 as record
import Q2 as display

def main():
    record.main()
    display.main()


if __name__ == "__main__":
    main()
这是Q1，这会造成问题：

import sys
from PySide2 import QtWidgets as qtw
from PySide2 import QtGui as qtg
from PySide2 import QtCore as qtc
from PySide2 import QtMultimedia as qtmm


class MainWindow(qtw.QMainWindow):
    def __init__(self):
        super().__init__()

        #Create Window layout with a sound widget
        soundboard = qtw.QWidget()
        soundboard.setLayout(qtw.QGridLayout())
        self.setCentralWidget(soundboard)
        sw = SoundWidget()
        soundboard.layout().addWidget(sw)

        #Window Dimensions
        self.setSizePolicy(qtw.QSizePolicy.Expanding, qtw.QSizePolicy.MinimumExpanding)

        # Code ends here
        self.show()


class SendOrderButton(qtw.QPushButton):
    button_stylesheet = 'background-color: blue; color: white;'

    def __init__(self):
        super().__init__('Send Order')
        self.setSizePolicy(qtw.QSizePolicy.Expanding, qtw.QSizePolicy.Expanding)
        self.setStyleSheet(self.button_stylesheet)
        #self.clicked.connect(qtc.QCoreApplication.instance().quit)

    def press_button(self):
        if self.isEnabled():
            self.setEnabled(False)
            self.setText('Send Order')
        else:
            self.setEnabled(True)
            self.setText('Sent')

class SoundWidget(qtw.QWidget):
    def __init__(self):
        super().__init__()
        self.setLayout(qtw.QGridLayout())

        #Send Order Button
        self.sendorder_button = SendOrderButton()
        self.sendorder_button.setShortcut(qtg.QKeySequence('Tab'))
        self.layout().addWidget(self.sendorder_button, 5, 0, 1, 2)
        self.sendorder_button.clicked.connect(qtc.QCoreApplication.instance().quit)

def main():
    app = qtw.QApplication(sys.argv)
    window = MainWindow()
    window.show()
    app.exec_()

if __name__ == '__main__':
    app = qtw.QApplication(sys.argv)
    window = MainWindow()
    window.show()
    sys.exit(app.exec_())
这是Q2.py，它具有第二个QApplication：

import sys
from PySide2.QtCore import (QAbstractTableModel, Slot)
from PySide2.QtWidgets import (QAction, QApplication, QMainWindow,QWidget)

class MainWindow(QMainWindow):
    def __init__(self, widget):
        QMainWindow.__init__(self)

        # Exit QAction
        exit_action = QAction("Exit", self)
        exit_action.setShortcut("Ctrl+Q")
        exit_action.triggered.connect(self.exit_app)

    @Slot()
    def exit_app(self, checked):
        sys.exit()

class CustomTableModel(QAbstractTableModel):
    def __init__(self, data=None):
        QAbstractTableModel.__init__(self)

class Widget(QWidget):
    def __init__(self):
        QWidget.__init__(self)

        # Getting the Model
        self.model = CustomTableModel()

def main():
    app2 = QApplication()
    widget = Widget()
    window2 = MainWindow(widget)
    window2.show()
    sys.exit(app2.exec_())

if __name__ == "__main__":
    app = QApplication()
    widget = Widget()
    window = MainWindow(widget)
    window.show()
    sys.exit(app.exec_())

我想创建一个QApplication，然后使用键盘快捷键退出它。然后，python脚本应调用另一个QApplication。我目前的问题是，当第二个...

Answer 1

如评论中所述，Qt应用程序只能并且应该具有QApplication（您可能不遵循此规则，但是不能保证它可以正常工作，因此您将不得不重新构造代码。