Displaying pop-up windows in Python (PyQt4)

Question

I need to know how to be able to make a dialog pop-up when a user clicks a button.

I\'m relatively new to both Python and PyQt/QtDesigner. I\'ve only been using in

Answer 1

So as you can see, I've connected the first button to a method named 'popup', which needs to be filled in with code to make my second window pop up. How do I go about doing this?

Pretty much the same way you do it for your main window (MyForm).

As usual, you write a wrapper class for your QtDesigner code for the second dialog (like you did with MyForm). Let's call it MyPopupDialog. Then in your popup method, you create an instance and then show your instance with either exec_() or show() depending whether you want a modal or modeless dialog. (If you are not familiar with Modal/Modeless concept, you might refer to the documentation.)

So the overall thing might look like this (with a couple of modifications):

# Necessary imports

class MyPopupDialog(QtGui.QDialog):
    def __init__(self, parent=None):
        # Regular init stuff...
        # and other things you might want


class MyForm(QtGui.QDialog):
    def __init__(self, parent=None):
        # Here, you should call the inherited class' init, which is QDialog
        QtGui.QDialog.__init__(self, parent)

        # Usual setup stuff
        self.ui = Ui_Dialog()
        self.ui.setupUi(self)

        # Use new style signal/slots
        self.ui.pushButton.clicked.connect(self.popup)     

        # Other things...

   def popup(self):
        self.dialog = MyPopupDialog()

        # For Modal dialogs
        self.dialog.exec_()

        # Or for modeless dialogs
        # self.dialog.show()

if __name__ == "__main__":
   app = QtGui.QApplication(sys.argv)
   myapp= MyForm()
   myapp.show()
   sys.exit(app.exec_())