Can I create a new style pyqt signal that isn't a field member of a class?
问题 So for the only way that I can see to create a style signal with PyQt4 is as follows: class MyCustomClass(QtCore.QThread): custom_signal = QtCore.pyqtSignal(str) My beef is if I declare that signal anywhere else, pyqt throws an error at me about how custom_signal doesn't have a connect() function. I would like to create a helper function to help remove the boilerplate/repeated code when I want to do something as simple as: starting a new thread, doing work in that thread, sending the result