Hi,
One easy/obvious improvement would be to delete all user's
identifiers and groups at once without iterating over them for
every user. Actually, iterating over the list of user_ids is not
necessary too I think.
code
session.query(Identifier).filter(Identifier.user_id.in_(user_ids)).delete()
session.query(User).filter(User.id.in_(user_ids)).delete()
/code
I'm not sure about deleting groups in your code - I suppose you
don't want to delete the actual group but only the association between
the user and the group, i.e. the record from the intermediate table.
But the idea is the same - .filter(blah-
blah.user_id.in_(user_ids)).delete()
An even better solution would be to set up proper cascade rules on
your relationships so all dependent items are deleted automatically
when a user is deleted. Then the method will be a one-liner:
code
session.query(User).filter(User.id.in_(user_ids)).delete()
/code
On Jun 1, 7:40 pm, Fabien Ribes fri...@gmail.com wrote:
Hi,
I have three tables (user, identifiers, groups) with 'many to one' and
'many to many' relationships. My code for deleting a user works
correctly but I would like to have it run faster. What improvements do
you suggest ?
code
def delete_list(self, user_ids):
Delete a list of users by id
session = self._get_orm_session()
for user_id in user_ids:
try:
user = session.query(User).filter(User.id ==
user_id).one()
if user:
# loop to free the user's identifiers
for ident in user.identifiers:
ident.user = None
# loop to remove the user from the groups
for group in user.groups:
del group
session.query(User).filter(User.id ==
user_id).delete()
except NoResultFound, ex:
raise SmartResourceNotFound(User not found +
str(ex))
session.commit()
/code
--
You received this message because you are subscribed to the Google Groups
sqlalchemy group.
To post to this group, send email to sqlalchemy@googlegroups.com.
To unsubscribe from this group, send email to
sqlalchemy+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/sqlalchemy?hl=en.
