PyQt4程序移植到PyQt5的时候出现问题啦！大家看看~

代码：
#!/usr/bin/python
# menubar.py
import sys
from PyQt5 import QtWidgets, QtCore,QtGui
class MainWindow(QtWidgets.QMainWindow):
def __init__(self, parent = None):
QtWidgets.QMainWindow.__init__(self)
self.resize(250, 150)
self.setWindowTitle('menubar')
exit = QtWidgets.QAction(QtGui.QIcon('icons/exit.png'), 'Exit', self)
exit.setShortcut('Ctrl+Q')
exit.setStatusTip('Exit application')
exit.connect(exit, QtCore.SIGNAL('triggered()'), QtGui.qApp,
QtCore.SLOT('quit()'))
self.statusBar()
menubar = self.menuBar()
file = menubar.addMenu('&File')
file.addAction(exit)
app = QtWidgets.QApplication(sys.argv)
main = MainWindow()
main.show()
sys.exit(app.exec_())
'''首先我们使用 QMainWindow 类的 menuBar()方法创建一个菜单栏。 然后使用 addMenu()
方法添加一个菜单。最后我们把动作对象（这里是 exit）添加到 file 菜单中。'''
错误：
Traceback (most recent call last):
File "E:/PyQt/Codes/menubar.py", line 20, in <module>
main = MainWindow()
File "E:/PyQt/Codes/menubar.py", line 13, in __init__
exit.connect(exit, QtCore.SIGNAL('triggered()'), QtGui.qApp,
AttributeError: 'QAction' object has no attribute 'connect'
>>>