This code works for building the menu at startup. It creates a submenu for each style found.
styleNames = QStyleFactory.keys()
for style in styleNames:
self.actionStyle = QAction(self)
self.actionStyle.setObjectName("actionStyle")
self.menu_Styles.addAction(self.actionStyle)
self.menu_Preferences.addAction(self.menu_Styles.menuAction())
self.menubar.addAction(self.menu_File.menuAction())
self.menubar.addAction(self.menu_Preferences.menuAction())
self.actionStyle.setText(style)
self.actionStyle.triggered.connect(self.changeStyle))
This is the changeStyle function. The purpose is to change the widget style based on the menu selection.
def changeStyle(self, styleName):
QApplication.setStyle(QStyleFactory.create(styleName))
The menu gets created properly but I cannot figure out how to pass the styleName string to the function.
I’ve done a similar thing with a combobox which allowed me to pass the string with self.combobox.activated[str].connect(self.changeStyle). Similar syntax with ‘triggered’ does not work.
The app starts and the menu look great but when I select one of the menu options I get an error about the number of arguments.
TypeError: changeStyle() takes exactly 2 arguments (1 given)
I am open to suggestions, advice, or comments.
Thanks in advance,
Cheers,
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