MenuBar单击事件

Question

我想在菜单栏单击（不是菜单项）后打开一个窗口。什么是错误的下面的代码：

self.menuHelp.triggered.connect(lambda : HelpWindow().exec_()) 

Answer 1

如果我理解正确，那么你想在菜单栏点击打开新窗口，trigger event与click事件不同，所以你需要实现自己的菜单栏，我有一个工作例如但不知道这是什么点:) 和一些代码，我从上面的答案，对不起懒惰。

import sys 
# This is bad, but was stealing code from above answer 
from PyQt4.QtCore import * 
from PyQt4.QtGui import * 

class HelpWindow(QtGui.QDialog): 
    def __init__(self): 
     super(HelpWindow, self).__init__() 
     self.setObjectName("self") 
     self.resize(500, 350) 
     layout = QtGui.QVBoxLayout() 
     textLbl = QtGui.QLabel("REally ?") 
     layout.addWidget(textLbl) 
     self.setLayout(layout) 

    def exec_(self): 
     super(HelpWindow, self).exec_() 
     return "Blaaa" 

class CustomMB(QMenuBar): 
    mbClick = pyqtSignal(str) 
    def __init__(self, parent=None): 
     super(CustomMB, self).__init__() 

    @pyqtSlot() 
    def mouseReleaseEvent(self, event): 
     self.mbClick.emit('clicked') 

class MyWindow(QMainWindow): 
    def __init__(self): 
     super(MyWindow, self).__init__() 
     self.main_menu = CustomMB() 
     self.main_menu.mbClick.connect(lambda : HelpWindow().exec_()) 
     self.setMenuBar(self.main_menu) 
     self.create_menu() 

    def create_menu(self): 
     menu2 = self.main_menu.addMenu('Menu 1') 

     Action1=QAction('Menu 1 0',self) 
     Action1.triggered.connect(self.action_1) 
     menu2.addAction(Action1) 

     Action2=QAction('Menu 1 1',self) 
     Action2.triggered.connect(self.action_2) 
     menu2.addAction(Action2) 

    def menu_1(self): 
     new=MyWindow() 
     new.setMinimumSize(320,160) 
     new.show() 

    def action_1(self): 
     print('Menu 1 0') 

    def action_2(self): 
     print('Menu 1 1') 


if __name__ == '__main__': 
    app=QApplication(sys.argv) 
    new=MyWindow() 
    new.show() 
    app.exec_()