CODEWITHSUNDEEP
pythonqtlambdapyqtqdialog
Pastrana27
Pastrana27

Reputation: 55

Pyqt Signal to open dialog lambda function error: Object Ui_dialog is not callable

My problem is that, when I put a Qaction triggered in pyqt to open a QDialog, I used to do with lambda functions (to pass parameters), but when I test the GUI, I opened for a first time the dialog, but, when I opened for a second time, the shell throws me this error

    Traceback (most recent call last):
  File "C:\path\to\file\launcher.py", line 51, in <lambda>
    self.ui.actionIngresar_Licencia.triggered.connect(lambda: self.Dlg_IngresarLicencia())
TypeError: 'Ui_dialogoLicencia' object is not callable

Ok, you can say me that I don pass any parameter, but if this error appears when I need to pass parameters, It will be so bad. When I put the function without lambda function, the gui works perfectly Here's the function with the dialog.

def Dlg_IngresarLicencia(self):
    self.Dlg_IngresarLicencia = Ui_dialogoLicencia()
    self.dialogo = QtGui.QDialog(parent=None)
    self.Dlg_IngresarLicencia.setupUi(self.dialogo)
    self.Dlg_IngresarLicencia.btn_ObtenerLicencia.clicked.connect(lambda: Componentes().clickObtenerLicencia())
    self.dialogo.show()

Thanks guys, I hope you can help me

Upvotes: 0

Views: 223

Answers (1)

The Compiler
The Compiler

Reputation: 11939

Your method is called the same as the attribute you're setting:

def Dlg_IngresarLicencia(self):
    self.Dlg_IngresarLicencia = Ui_dialogoLicencia()

Because of that, after the first call you override the method with a Ui_dialogoLicencia instance, which is not callable.

Upvotes: 1

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