Whats the correct way of showing different window after action/button trigger in PyQt4 using OOP?
Do I just make a different class for the window and put its instance to a method ? Like so:
import sys
from PyQt4 import QtGui, QtCore, uic
class Window(QtGui.QMainWindow):
def __init__(self):
super(Window, self).__init__()
uic.loadUi("ui/main.ui", self)
self.show()
self.settingsAction.triggered.connect(lambda: self.settingsTrigger())
def settingsTrigger(self):
self.settingsDialog = Settings()
class Settings(QtGui.QMainWindow):
def __init__(self):
super(Settings, self).__init__()
uic.loadUi("ui/settings.ui", self)
self.show()
app = QtGui.QApplication(sys.argv)
GUI = Window()
sys.exit(app.exec_())
With settings being the second window. The above code works but I highly doubt that this is how its supposed to be done.
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This would be a great question for /r/learnpython. I know OOP, but no expert. Your code looks fine to me though in all honesty.
-Harrison 9 years ago
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Don't need expert opinion I'm fine with knowing its nothing too bad to do it this way. Thanks :)
-FlyingEagle 9 years ago
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