there are many ways of solving this problem. please see the below thread for more details where similar problem was discussed : http://groups.google.com/group/django-users/browse_thread/thread/3a9960139307aaec/77b712ee9c08795b#77b712ee9c08795b
Another method which you can use is ==================================== for respective model override the save as follows class Model(): #this code syntax may not be proper def save(self): super(self) #at this point it will save and you will know the id self.account_number = "X"+self.id self.save() The syntax may be wrong as i have did not do it earlier The core idea is to override the save and then call its parent to save the object,after saving using that object id form account number and save it again On Feb 13, 12:54 pm, gganesh <ganesh....@gmail.com> wrote: > hello , > i will be glad to find your help in the task mentioned below , > i need to create a account_number column which is a concat of id > column and with some variable ,for example, if my id column is 1 ( id > =1) and account_number should be X1(account_number=X1) and if my id is > 2 the account_number must be X2 and so on > thanks --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---