PyQt : Signal like this

[prashant]

I am creating <QAction> in a loop and adding it to menu.

for fileName in fileNames:
    action = QtGui.QAction(QtGui.QIcon(iconFile), fileName, menu)
    menu.addAction(action)

for fileName in fileNames:
    action = QtGui.QAction(QtGui.QIcon(iconFile), fileName, menu)
    menu.addAction(action)

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I need to connect this action to a python function in such a way that I should get <text> of QAction <fileName> or instance of QAction object itself.

For example If action name is required

def contextMenuAction(self, actionName):
        print actionName

def contextMenuAction(self, actionName):
        print actionName

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If action object is required

def contextMenuAction(self, actionObject):
        myData = actionObject.data()

def contextMenuAction(self, actionObject):
        myData = actionObject.data()

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How do I setup action and slot? Logically it would be like this:

self.connect(action, QtCore.SIGNAL("triggered()"), self, QtCore.SLOT("contextMenuAction(str)"), fileName)
self.connect(action, QtCore.SIGNAL("triggered()"), self, QtCore.SLOT("contextMenuAction(object)"), action)

self.connect(action, QtCore.SIGNAL("triggered()"), self, QtCore.SLOT("contextMenuAction(str)"), fileName)
self.connect(action, QtCore.SIGNAL("triggered()"), self, QtCore.SLOT("contextMenuAction(object)"), action)

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