Re: [sqlalchemy] Updating a one-to-many relationship
I think you may be confused about the relationship properties you have
here. As far as I can tell, a Creator can have many companies, but
each Company has only one creator, correct? So Company.creator should
only ever be an instance of Creator (or None), whereas
Creator.companies should be a list.
In your __repr__ example:
class Creator(Base):
def __repr__(self):
return '%s' % self.creator
class Company(Base):
def __repr__(self):
return '%s, created by %s' % (self.company, self.creator[0])
Why are you using self.creator[0] here? self.creator is not a list,
it should either be an instance of Creator, or None.
Overriding __repr__ is also a good way to make debugging difficult.
For example, if you had a list of Creator instances and you printed
them at the python prompt, it would just look like a list of strings.
When I want extra information from __repr__, I normally write it
something like this:
def __repr__(self):
classname = type(self).__name__
return '%s name=%r' % (classname, self.name)
In your second example:
a=session.query(Creator).first()
a[0].companies
a.companies
Query.first() returns a single value, not a list. So typing a[0]
doesn't make any sense.
Please try to create a self-contained script that demonstrates your
problem. Here is a good example:
https://groups.google.com/d/msg/sqlalchemy/jQtIRJXVfH8/LgwX-bomEIQJ
Thanks,
Simon
On Wed, Aug 14, 2013 at 2:45 AM, csdr...@gmail.com wrote:
I'm afraid there are still some bugs in here that hopefully you can help
with.
class Creator(Base):
__tablename__ = creators
id = Column(Integer, primary_key = True)
company_id = Column(Integer, ForeignKey('companies.id'))
creator = Column(String(100), nullable=False, unique=True)
def __init__(self, creator):
self.creator = creator
def __repr__(self):
return '%s' % self.creator # otherwise returns a single entry list
for some reason (e.g. would display [user])
class Company(Base):
__tablename__ = companies
id = Column(Integer, primary_key = True)
company = Column(String(100), unique=True, nullable=False) #might want
to revise string sizes at some point
creator = relationship(Creator, backref=companies, cascade=all)
def __init__(self, company, creator):
self.company = company
#self.creator.append(Creator(creator))
existing_creator =
session.query(Creator).filter_by(creator=creator).first()
#self.creator.append(existing_creator or Creator(creator))
if existing_creator:
print True
self.creator.append(existing_creator)
else:
self.creator.append(Creator(creator))
def __repr__(self):
return '%s, created by %s' % (self.company, self.creator[0])
1) Weird __repr__ error:
class Creator(Base):
def __repr__(self):
return '%s' % self.creator
class Company(Base):
def __repr__(self):
return '%s, created by %s' % (self.company, self.creator[0])
c=Company(Company1, mike)
session.add(c)
c=Company(Company2, mike)
True
session.add(c)
c=Company(Company3, john)
session.add(c)
c=Company(Company4, mike)
True
session.add(c)
session.query(Company).all()
[Traceback (most recent call last):
File stdin, line 1, in module
File stdin, line 17, in __repr__
However, if I divide the query lines among every add() statement, there is
no __repr__ error.
c=Company(Company1, mike)
session.add(c)
session.query(Company).all()
[Company1, created by mike]
c=Company(Company2, mike)
True
session.add(c)
session.query(Company).all()
[Company1, created by mike, Company2, created by mike]
c=Company(Company3, john)
session.add(c)
session.query(Company).all()
[Company1, created by mike, Company2, created by mike, Company3, created by
john]
c=Company(Company4, mike)
True
session.add(c)
session.query(Company).all()
[Company1, created by mike, Company2, created by mike, Company3, created by
john, Company4, created by mike]
2) Creator.companies only shows the most recently added company:
session.query(Company).all()
[Company1, created by mike, Company2, created by mike, Company3, created by
john, Company4, created by mike]
session.query(Creator).all()
[mike, john]
a=session.query(Creator).first()
a[0].companies
a.companies
Company4, created by mike
3) Weird Company.creator error:
session.query(Company).all()
[Company1, created by mike, Company2, created by mike, Company3, created by
john, Company4, created by mike]
session.query(Company.creator).all()
[(False,), (False,), (False,), (False,), (True,), (False,), (False,),
(True,)]
a=session.query(Company).first()
a.creator
[mike]
Anyone have any ideas?
--
You received this message because you are subscribed to the Google Groups
sqlalchemy group.
To unsubscribe from this group and stop receiving emails from it, send an
email to sqlalchemy+unsubscr...@googlegroups.com.
To
Re: [sqlalchemy] Cross-schema foreign keys reflection
On Tuesday, August 13, 2013 7:43:54 PM UTC-4, Michael Bayer wrote:
cross-schema reflection is supported on PG but has caveats, see
http://docs.sqlalchemy.org/en/rel_0_8/dialects/postgresql.html#remote-cross-schema-table-introspectionfor
a discussion of recommended usage patterns.
Thanks for pointing out that page. I also needed to add the schema to the
foreign key declarations.
eg Column(Integer, ForeignKey('otherschema.othertable.id')) instead of
Column(Integer, ForeignKey('othertable.id'))
-- Jason
--
You received this message because you are subscribed to the Google Groups
sqlalchemy group.
To unsubscribe from this group and stop receiving emails from it, send an email
to sqlalchemy+unsubscr...@googlegroups.com.
To post to this group, send email to sqlalchemy@googlegroups.com.
Visit this group at http://groups.google.com/group/sqlalchemy.
For more options, visit https://groups.google.com/groups/opt_out.
[sqlalchemy] many-to-one relationship with intermediate table non equijoin
Hello all
Tried for hours to figure out the various relationship() options with no
luck.
Consider:
class Enrolment(base):
__tablename__ = 'enrolment'
person_id = Column(String, primary_key=True)
group_id= Column(String, primary_key=True)
enrol_date = Column(Date, primary_key=True)
level_id= Column(String, nullable=False)
next_date = Column(Date)
def __repr__(self):
return 'Enrol(%s, %s, %s, %s)' % (self.person_id,
self.enrol_date,
self.group_id,
self.level_id)
class RosterLine(base):
__tablename__ = 'roster_line'
line_id = Column(String, primary_key=True)
group_id= Column(String, nullable=False)
class Timesheet(base):
__tablename__ = 'timesheet'
id = Column(Integer, primary_key=True)
person_id = Column(String, nullable=False)
line_id = Column(String, nullable=False)
date= Column(Date, nullable=False)
enrolment = relationship(Enrolment,
primaryjoin=lambda:(
(Timesheet.person_id ==
foreign(Enrolment.person_id))
(Timesheet.date = Enrolment.enrol_date)
((Timesheet.date Enrolment.next_date)
| (Enrolment.next_date == None))
# (Timesheet.line_id ==
RosterLine.line_id)
# (RosterLine.group_id ==
Enrolment.group_id)
),
# uselist=False,
viewonly=True)
The relationship as it stands works correctly but I can't figure out the
magic words to introduce the intermediate join to RosterLine.
The relationship should issue SQL like:
select E.*
fromroster_line L,
enrolment E
where L.line_id = 'work'
and L.group_id = E.group_id
and E.person_id = 'bob'
and E.enrol_date = '2012-03-04'
and (E.next_date '2012-03-04'
or E.next_date is null)
Eternally grateful for any help.
Thanks.
--
You received this message because you are subscribed to the Google Groups
sqlalchemy group.
To unsubscribe from this group and stop receiving emails from it, send an email
to sqlalchemy+unsubscr...@googlegroups.com.
To post to this group, send email to sqlalchemy@googlegroups.com.
Visit this group at http://groups.google.com/group/sqlalchemy.
For more options, visit https://groups.google.com/groups/opt_out.
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
from datetime import date
base = declarative_base()
class Enrolment(base):
__tablename__ = 'enrolment'
person_id = Column(String, primary_key=True)
group_id= Column(String, primary_key=True)
enrol_date = Column(Date, primary_key=True)
level_id= Column(String, nullable=False)
next_date = Column(Date)
def __repr__(self):
return 'Enrol(%s, %s, %s, %s)' % (self.person_id,
self.enrol_date,
self.group_id,
self.level_id)
class RosterLine(base):
__tablename__ = 'roster_line'
line_id = Column(String, primary_key=True)
group_id= Column(String, nullable=False)
class Timesheet(base):
__tablename__ = 'timesheet'
id = Column(Integer, primary_key=True)
person_id = Column(String, nullable=False)
line_id = Column(String, nullable=False)
date= Column(Date, nullable=False)
enrolment = relationship(Enrolment,
primaryjoin=lambda:(
(Timesheet.person_id == foreign(Enrolment.person_id))
(Timesheet.date = Enrolment.enrol_date)
((Timesheet.date Enrolment.next_date)
| (Enrolment.next_date == None))
# (Timesheet.line_id == RosterLine.line_id)
# (RosterLine.group_id == Enrolment.group_id)
),
#uselist=False,
viewonly=True)
e = create_engine('sqlite://', echo=True)
base.metadata.create_all(e)
db = Session(e)
db.add(RosterLine(line_id='work', group_id='staff'))
db.add(RosterLine(line_id='etc', group_id='manager'))
db.add(Enrolment(person_id='bob',
group_id='staff',
level_id='normal',
enrol_date=date(2010,1,1),
next_date=date(2011,1,1)))
db.add(Enrolment(person_id='bob',
group_id='staff',
level_id='better',
enrol_date=date(2011,1,1)))
Re: [sqlalchemy] many-to-one relationship with intermediate table non equijoin
create a non primary mapper to a select() that's against the Enrolment table joined to RosterLine (i.e. mapper(myselect.alias(), non_primary=True), then construct a relationship() to that mapper (viewonly=True of course). at some point I should add an example of this technique, it's just the easiest way to deal with the relationship-across-any-number-of-tables use case. On Aug 14, 2013, at 2:30 PM, avdd adr...@gmail.com wrote: Hello all Tried for hours to figure out the various relationship() options with no luck. Consider: class Enrolment(base): __tablename__ = 'enrolment' person_id = Column(String, primary_key=True) group_id= Column(String, primary_key=True) enrol_date = Column(Date, primary_key=True) level_id= Column(String, nullable=False) next_date = Column(Date) def __repr__(self): return 'Enrol(%s, %s, %s, %s)' % (self.person_id, self.enrol_date, self.group_id, self.level_id) class RosterLine(base): __tablename__ = 'roster_line' line_id = Column(String, primary_key=True) group_id= Column(String, nullable=False) class Timesheet(base): __tablename__ = 'timesheet' id = Column(Integer, primary_key=True) person_id = Column(String, nullable=False) line_id = Column(String, nullable=False) date= Column(Date, nullable=False) enrolment = relationship(Enrolment, primaryjoin=lambda:( (Timesheet.person_id == foreign(Enrolment.person_id)) (Timesheet.date = Enrolment.enrol_date) ((Timesheet.date Enrolment.next_date) | (Enrolment.next_date == None)) # (Timesheet.line_id == RosterLine.line_id) # (RosterLine.group_id == Enrolment.group_id) ), # uselist=False, viewonly=True) The relationship as it stands works correctly but I can't figure out the magic words to introduce the intermediate join to RosterLine. The relationship should issue SQL like: select E.* fromroster_line L, enrolment E where L.line_id = 'work' and L.group_id = E.group_id and E.person_id = 'bob' and E.enrol_date = '2012-03-04' and (E.next_date '2012-03-04' or E.next_date is null) Eternally grateful for any help. Thanks. -- You received this message because you are subscribed to the Google Groups sqlalchemy group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/groups/opt_out. myrelation.py signature.asc Description: Message signed with OpenPGP using GPGMail
Re: [sqlalchemy] Updating a one-to-many relationship
Simon, your idea about putting together a script is a good one. Please see
the attached. I think all these errors are related but I'm scratching my
head about what the problem is.
The reason I use self.creator[0] versus self.creator is for aesthetics.
And, to your point about creator not being a list, SQLAlchemy is treating
creator as a list-like object. Since any company can only have one creator,
I wasn't concerned about indexing creator[0].
a=session.query(Company).first()
a
Company1, created by mike
rather than
a=session.query(Company).first()
a
Company1, created by [mike]
a.creator
[mike]
More info on creator:
type(a.creator)
class 'sqlalchemy.orm.collections.InstrumentedList'
type(a.creator[0])
class '__main__.Creator'
Chris
On Wednesday, August 14, 2013 6:18:51 AM UTC-4, Simon King wrote:
I think you may be confused about the relationship properties you have
here. As far as I can tell, a Creator can have many companies, but
each Company has only one creator, correct? So Company.creator should
only ever be an instance of Creator (or None), whereas
Creator.companies should be a list.
In your __repr__ example:
class Creator(Base):
def __repr__(self):
return '%s' % self.creator
class Company(Base):
def __repr__(self):
return '%s, created by %s' % (self.company, self.creator[0])
Why are you using self.creator[0] here? self.creator is not a list,
it should either be an instance of Creator, or None.
Overriding __repr__ is also a good way to make debugging difficult.
For example, if you had a list of Creator instances and you printed
them at the python prompt, it would just look like a list of strings.
When I want extra information from __repr__, I normally write it
something like this:
def __repr__(self):
classname = type(self).__name__
return '%s name=%r' % (classname, self.name)
In your second example:
a=session.query(Creator).first()
a[0].companies
a.companies
Query.first() returns a single value, not a list. So typing a[0]
doesn't make any sense.
Please try to create a self-contained script that demonstrates your
problem. Here is a good example:
https://groups.google.com/d/msg/sqlalchemy/jQtIRJXVfH8/LgwX-bomEIQJ
Thanks,
Simon
On Wed, Aug 14, 2013 at 2:45 AM, csd...@gmail.com javascript: wrote:
I'm afraid there are still some bugs in here that hopefully you can help
with.
class Creator(Base):
__tablename__ = creators
id = Column(Integer, primary_key = True)
company_id = Column(Integer, ForeignKey('companies.id'))
creator = Column(String(100), nullable=False, unique=True)
def __init__(self, creator):
self.creator = creator
def __repr__(self):
return '%s' % self.creator # otherwise returns a single entry
list
for some reason (e.g. would display [user])
class Company(Base):
__tablename__ = companies
id = Column(Integer, primary_key = True)
company = Column(String(100), unique=True, nullable=False) #might
want
to revise string sizes at some point
creator = relationship(Creator, backref=companies,
cascade=all)
def __init__(self, company, creator):
self.company = company
#self.creator.append(Creator(creator))
existing_creator =
session.query(Creator).filter_by(creator=creator).first()
#self.creator.append(existing_creator or Creator(creator))
if existing_creator:
print True
self.creator.append(existing_creator)
else:
self.creator.append(Creator(creator))
def __repr__(self):
return '%s, created by %s' % (self.company, self.creator[0])
1) Weird __repr__ error:
class Creator(Base):
def __repr__(self):
return '%s' % self.creator
class Company(Base):
def __repr__(self):
return '%s, created by %s' % (self.company, self.creator[0])
c=Company(Company1, mike)
session.add(c)
c=Company(Company2, mike)
True
session.add(c)
c=Company(Company3, john)
session.add(c)
c=Company(Company4, mike)
True
session.add(c)
session.query(Company).all()
[Traceback (most recent call last):
File stdin, line 1, in module
File stdin, line 17, in __repr__
However, if I divide the query lines among every add() statement, there
is
no __repr__ error.
c=Company(Company1, mike)
session.add(c)
session.query(Company).all()
[Company1, created by mike]
c=Company(Company2, mike)
True
session.add(c)
session.query(Company).all()
[Company1, created by mike, Company2, created by mike]
c=Company(Company3, john)
session.add(c)
session.query(Company).all()
[Company1, created by mike, Company2, created by mike, Company3, created
by
john]
c=Company(Company4, mike)
True
session.add(c)
[sqlalchemy] Under what circumstances will a `inspect(myobject).attrs.myattribute.history` be a sequence of more than one item ?
As previously discussed, i'm using an object's history to log changes to the database within an application. http://docs.sqlalchemy.org/en/rel_0_8/orm/session.html?highlight=history#sqlalchemy.orm.attributes.History Each tuple member is an iterable sequence I'm trying to figure out when I would expect that sequence to have more than 1 item in it. My guess is that for object relationships, it might contain a list of more than one element -- but for column data it would only be a single element. is that correct or am I off ? -- You received this message because you are subscribed to the Google Groups sqlalchemy group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/groups/opt_out.
[sqlalchemy] TypeDecorator to store bool as ENUM('N', 'Y')?
I'm working with an existing MySQL schema that has lots of columns of
type ENUM('N', 'Y'). I'd like to deal with them as real booleans on
the python side.
I have a simple TypeDecorator which almost works (I think):
class YNBoolean(sqlalchemy.types.TypeDecorator):
impl = mysql.ENUM('N', 'Y', charset='ascii')
def process_bind_param(self, value, dialect):
if value is None:
return value
return 'Y' if value else 'N'
def process_result_value(self, value, dialect):
if value is None:
return None
return value == 'Y'
The one problem I've discovered with this is that
session.query(MyTable).filter(MyTable.ynbool)
produces a query like
SELECT ... FROM MyTable WHERE MyTable.ynbool;
What I really want is
SELECT ... FROM MyTable WHERE MyTable.ynbool = 'Y';
(If I do .filter(MyTable.ynbool == True) that does work as desired/expected.)
Is there a way to customize how my column gets compiled when used
in an expression in a boolean context? (If not, I can live with it
as is. I'll just get surprised once in awhile when I forget that
treating the column as a boolean in expressions won't work.)
Thank you for any help.
Jeff
--
You received this message because you are subscribed to the Google Groups
sqlalchemy group.
To unsubscribe from this group and stop receiving emails from it, send an email
to sqlalchemy+unsubscr...@googlegroups.com.
To post to this group, send email to sqlalchemy@googlegroups.com.
Visit this group at http://groups.google.com/group/sqlalchemy.
For more options, visit https://groups.google.com/groups/opt_out.
[sqlalchemy] Re: many-to-one relationship with intermediate table non equijoin
Thanks for the quick response! After much fiddling I got it working using alias(), foreign() and corresponding_column(). It seems to get the right results. Is this the simplest, right approach? joined = Enrolment.__table__.join(RosterLine, Enrolment.group_id==RosterLine.group_id).alias() cc = joined.corresponding_column secmapper = mapper(Enrolment, joined, non_primary=True) # ... enrolment = relationship(secmapper, primaryjoin=lambda:( (Timesheet.line_id == foreign(cc(RosterLine.line_id))) (Timesheet.person_id == cc(Enrolment.person_id)) (Timesheet.date = cc(Enrolment.enrol_date)) ((Timesheet.date cc(Enrolment.next_date)) | (cc(Enrolment.next_date) == None)) ), uselist=False, viewonly=True) On Thursday, 15 August 2013 04:30:12 UTC+10, avdd wrote: Hello all Tried for hours to figure out the various relationship() options with no luck. Consider: class Enrolment(base): __tablename__ = 'enrolment' person_id = Column(String, primary_key=True) group_id= Column(String, primary_key=True) enrol_date = Column(Date, primary_key=True) level_id= Column(String, nullable=False) next_date = Column(Date) def __repr__(self): return 'Enrol(%s, %s, %s, %s)' % (self.person_id, self.enrol_date, self.group_id, self.level_id) class RosterLine(base): __tablename__ = 'roster_line' line_id = Column(String, primary_key=True) group_id= Column(String, nullable=False) class Timesheet(base): __tablename__ = 'timesheet' id = Column(Integer, primary_key=True) person_id = Column(String, nullable=False) line_id = Column(String, nullable=False) date= Column(Date, nullable=False) enrolment = relationship(Enrolment, primaryjoin=lambda:( (Timesheet.person_id == foreign(Enrolment.person_id)) (Timesheet.date = Enrolment.enrol_date) ((Timesheet.date Enrolment.next_date) | (Enrolment.next_date == None)) # (Timesheet.line_id == RosterLine.line_id) # (RosterLine.group_id == Enrolment.group_id) ), # uselist=False, viewonly=True) The relationship as it stands works correctly but I can't figure out the magic words to introduce the intermediate join to RosterLine. The relationship should issue SQL like: select E.* fromroster_line L, enrolment E where L.line_id = 'work' and L.group_id = E.group_id and E.person_id = 'bob' and E.enrol_date = '2012-03-04' and (E.next_date '2012-03-04' or E.next_date is null) Eternally grateful for any help. Thanks. -- You received this message because you are subscribed to the Google Groups sqlalchemy group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/groups/opt_out.
[sqlalchemy] how to tell which columns are deferred in orm ?
i'm trying to generate a list of non-deffered columns from an object referencing this example: class Book(Base): __tablename__ = 'book' book_id = Column(Integer, primary_key=True) title = Column(String(200), nullable=False) summary = Column(String(2000)) excerpt = deferred(Column(Text)) photo = deferred(Column(Binary)) how can I tell , when dealing with an instance of `Book()`, which columns are deferred or not ? i tried inspecting virtually everything in: class_mapper(Book)) and looked over these multiple times: class_mapper(Book).mapped_table) class_mapper(Book).mapped_table.c) but couldn't seem to find anything that works. -- You received this message because you are subscribed to the Google Groups sqlalchemy group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/groups/opt_out.
[sqlalchemy] Re: DeferredReflection and mapper
On Wednesday, August 14, 2013 12:10:12 AM UTC-5, Lukasz Szybalski wrote:
Hello,
How do I go from class like defeinition to below with mapper.
The docs in 0.8 say I can use:
from sqlalchemy.ext.declarative import DeferredReflectionBase =
declarative_base()
class MyClass(DeferredReflection, Base):
__tablename__ = 'mytable'
but how do I do below with DefferedReflection
--
from sqlalchemy import Table
from sqlalchemy.orm import mapper, relation
class Recall(object):
def __init__(self, **kw):
automatically mapping attributes
for key, value in kw.iteritems():
setattr(self, key, value)
recall_table = Table('recall_db', metadata,
autoload=True,autoload_with=engine)
mapper(Recall, recall_table,primary_key=[recall_table.c.RECORD_ID])
I'm using pyramid, and I want to autoload tables in models.py which does
not have the engine bound yet. I will bind in in main function with
DeferredReflection.prepare(engine)
Never mind:
I followed the documentation:
In models.py I did
from sqlalchemy.ext.declarative import DeferredReflection
class Recall(DeferredReflection, Base):
__tablename__ = 'recall_db'
in __init__.py I did
from sqlalchemy.ext.declarative import DeferredReflection
then under
Base.metadata.bind = engine
add
DeferredReflection.prepare(engine)
I can access the table with
Recall.__table__
and mapper with
Recall.__mapper__
I presume this does the same as :
recall_table = Table('recall_db', metadata, autoload=True,autoload_with=
engine)
class Recall(object):
def __init__(self, **kw):
automatically mapping attributes
for key, value in kw.iteritems():
setattr(self, key, value)
mapper(Recall, recall_table,primary_key=[recall_table.c.RECORD_ID])
Thanks
Lucas
--
You received this message because you are subscribed to the Google Groups
sqlalchemy group.
To unsubscribe from this group and stop receiving emails from it, send an email
to sqlalchemy+unsubscr...@googlegroups.com.
To post to this group, send email to sqlalchemy@googlegroups.com.
Visit this group at http://groups.google.com/group/sqlalchemy.
For more options, visit https://groups.google.com/groups/opt_out.
Re: [sqlalchemy] Under what circumstances will a `inspect(myobject).attrs.myattribute.history` be a sequence of more than one item ?
yes it will always be one element for a scalar reference, a collection for collections. the ORM internally treats everything like a collection, kind of another artifact that probably wouldnt have been the case if this API were written for end-users originally... On Aug 14, 2013, at 9:20 PM, Jonathan Vanasco jvana...@gmail.com wrote: As previously discussed, i'm using an object's history to log changes to the database within an application. http://docs.sqlalchemy.org/en/rel_0_8/orm/session.html?highlight=history#sqlalchemy.orm.attributes.History Each tuple member is an iterable sequence I'm trying to figure out when I would expect that sequence to have more than 1 item in it. My guess is that for object relationships, it might contain a list of more than one element -- but for column data it would only be a single element. is that correct or am I off ? -- You received this message because you are subscribed to the Google Groups sqlalchemy group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/groups/opt_out. signature.asc Description: Message signed with OpenPGP using GPGMail
