esCUITValida

#!/usr/bin/env python
# -*- coding: utf-8 -*-
 
# <Aplicacion en Python para verificar la valides de la CUIT.>
# Copyright (C) 2012 Alejandro Alvarez <[email protected]>
#
# This program is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program. If not, see <http://www.gnu.org/licenses/>.
 
# http://www.codigopython.com.ar <[email protected]>
 
def esCUITValida(cuit):
    """
Funcion destinada a la validacion de CUIT
"""
    # Convertimos el valor a una cadena
    cuit = str(cuit)
    # Aca se ve si es de la forma de 13 caracteres
    # con los guiones
    if len(cuit) == 13 and cuit[2] == '-' and cuit[11] == '-':
        # Removemos los guiones para trabajar
        cuit = cuit.replace('-', '')
    # Si no tiene 11 caracteres lo descartamos
    elif len(cuit) != 11:
        return False, cuit
    # Solo resta analizar si todos los caracteres son numeros
    if not cuit.isdigit():
        return False, cuit
    # Despues de estas validaciones podemos afirmar
    # que contamos con 11 numeros
    base = [5, 4, 3, 2, 7, 6, 5, 4, 3, 2]
    aux = 0
    for i in xrange(10):
        aux += int(cuit[i]) * base[i]
    aux = 11 - (aux % 11)
    if aux == 11:
        aux = 0
    elif aux == 10:
        aux = 9
    if int(cuit[10]) == aux:
        return True, cuit
    else:
        return False, cuit