Action Button in MenuBar

[LT-K101]

Hi,
I have created a QMenuBar with QMenu object "File" with a sub Qaction button "Add New".
I want the Qaction button to open a new window. Below is my code, please can anyone confirm if I'm doing the right thing? Thanks.

self.ui.actionAdd_New.triggered.connect(self.window1)


 def window1(self):
     self.ui.w = Window1()
     self.ui.w.show()
       

[jsulm]

@LT-K101 So, unlike you wrote at the beginning you want to show another page when clicking one of the menus (not another window).
So, then let go step by step:

1. window1() is called, right?
2. What happens then? Nothing? Or is wrong page shown?
3. What if you use setCurrentIndex instead of setCurrentWidget?

[jsulm]

@LT-K101 said in Action Button in MenuBar:

please can anyone confirm if I'm doing the right thing?

Does it work?
I'm just not sure why you store Window1 instance inside self.ui?

[LT-K101]

@jsulm It does not work though, I was trying my hands on it but I'm stacked. All what I see online is below which works fine.

self.ui.actionClose.triggered.connect(qApp.exit)

[jsulm]

@LT-K101 Is window1 called?

[LT-K101]

@jsulm You mean calling the window1 function when the Qaction Button is clicked?

[jsulm]

@LT-K101 Yes, is actionAdd_New triggered, so it calls window1?
You can simply print something inside window1 to see whether it is called...

[LT-K101]

@jsulm When I try to print it works. Displaying a new window is the problem now.

[jsulm]

@LT-K101 How is Window1 implemented? Can you show its code?

[LT-K101]

@jsulm It is a QMainWindow with QStackedWidget pages in it. All what I want to do is show the first QStackedWidget page when i trigger the Qmenu action Button.

from PyQt5 import QtCore, QtGui, QtWidgets
from PyQt5.QtWidgets import QMainWindow, QAction,qApp,QApplication

class AdminForm(QtWidgets.QMainWindow):

    def __init__(self, mainMenu):
        super(AdminForm, self).__init__()
        self.mainMenu = mainMenu

        self.ui = Ui_MainWindow()
        self.ui.setupUi(self)

        self.ui.actionClose.triggered.connect(qApp.exit)

        self.ui.actionAdd_New_Employee.triggered.connect(self.window1)


    def window1(self):
        self.ui.stackedWidget.setCurrentWidget(self.ui.First_page)

[jsulm]

@LT-K101 said in Action Button in MenuBar:

AdminForm

I asked about Window1 which you're trying to show in your window1! Why do you show AdminForm? And why do you have a completely different implementation of window1 in AdminForm than the one you posted before? This is really confusing...

[LT-K101]

@jsulm what I'm showing is the class that controls the button actions with their define functions that can show the QStackedWidget pages .Initially I showed .hide() and .show() which I think is for showing QMainWindows, I stand to be corrected though.

[LT-K101]

@jsulm This is the GUI of my program, when I click on Add New Staff a pages should open. I'm really confused about how to implement this.

[jsulm]

@LT-K101 So, unlike you wrote at the beginning you want to show another page when clicking one of the menus (not another window).
So, then let go step by step:

1. window1() is called, right?
2. What happens then? Nothing? Or is wrong page shown?
3. What if you use setCurrentIndex instead of setCurrentWidget?

[LT-K101]

@jsulm said in Action Button in MenuBar:

setCurrentIndex

Thanks a lot I tried the setCurrentIndex method and it worked! like magic.