PyQT connect problem

I have this code:

Code:

---

def CreateActions(self):
    global openAct
    openAct = QtGui.QAction("Open", self)
    openAct.setShortcut("Ctrl+O")
    self.connect(openAct, QtCore.SIGNAL("triggered()"), self, QtCore.SLOT("self.Open(self)"))
def Open(self):
    print "test\n"

---

When I click the open button in my menu, nothing happens. When the code runs there are no errors. What is the problem?

I have this now:

Code:

---

def CreateActions(self):
        global openAct
        openAct = QtGui.QAction("Open", self)
        openAct.setShortcut("Ctrl+O")
        #openAct.triggered.connect(self.Open())
        QtCore.QObject.connect(openAct, QtCore.SIGNAL("triggered()"), self, self.Open())
def CreateMenus(self):
        menubar = self.menuBar()
        fileMenu = menubar.addMenu("&File")
        fileMenu.addAction(openAct)
def Open(self):
        print "test\n"
        fileName = QtGui.QFileDialog. \
            getOpenFileName(self, "Open File", 
                            QtCore.QDir.currentPath(), 
                            "NetDMF Files (*.xml *.xmf *.xmn)")

---

The "test" is printed, but the Open() method is executed as soon as I run the code. The program does not wait for me to click the Open action.

If I have this code:

Code:

---

def CreateActions(self):
        global openAct
        openAct = QtGui.QAction("Open", self)
        openAct.setShortcut("Ctrl+O")
        #openAct.triggered.connect(self.Open())
        QtCore.QObject.connect(openAct, QtCore.SIGNAL("triggered()"), self, self.Open())
def CreateMenus(self):
        menubar = self.menuBar()
        fileMenu = menubar.addMenu("&File")
        fileMenu.addAction(openAct)
def Open(self):
        print "test\n"

---

I get this error:
TypeError: argument 4 of QObject.connect() has an invalid type