PyQt sending parameter to slot when connecting to a signal

python qt4 pyqt
johannix picture johannix · Jun 2, 2009 · Viewed 32.1k times · Source

I have a taskbar menu that when clicked is connected to a slot that gets the trigger event. Now the problem is that I want to know which menu item was clicked, but I don't know how to send that information to the function connected to. Here is the used to connect the action to the function:

QtCore.QObject.connect(menuAction, 'triggered()', menuClickedFunc)

I know that some events return a value, but triggered() doesn't. So how do I make this happen? Do I have to make my own signal?

Answer

Eli Bendersky picture Eli Bendersky · Jun 2, 2009

Use a lambda

Here's an example from the PyQt book:

self.connect(button3, SIGNAL("clicked()"),
    lambda who="Three": self.anyButton(who))

By the way, you can also use functools.partial, but I find the lambda method simpler and clearer.

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