Re: [sqlalchemy] Can't locate strategy for ... (('lazy', 'joined'),)
On 5/21/15 3:56 PM, Russ wrote: nope. I'd need a complete, self-contained and succinct example I can run, thanks Ok, thanks. This is a beefy one so that will be extremely tricky to extract. I had hoped that the combo of lazy+joined would have been a clear indicator since they are opposite loading strategies. the word lazy there is the name of the field. It is the internal equivalent to the lazy argument on relationship: http://docs.sqlalchemy.org/en/rel_1_0/orm/relationship_api.html?highlight=relationship#sqlalchemy.orm.relationship.params.lazy that said, here is exactly what will cause your error. Any of the attribute names defined_items, child_product or number are not in fact bound to a relationship(), and instead refer to a Column-mapped attribute. Looks a lot like number here is a Column. Is that the case? You can't call joinedload for a column attribute. -- 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/d/optout.
Re: [sqlalchemy] relationsip(): In Parent or Child Class?
On 5/21/15 4:32 PM, Rich Shepard wrote: There are a number of many-to-one table/class relationships in the application. In the many class I use ForeignKey() to relate that column to the appropriate 'one' class and column. Reading the ORM tutorial tells me that the relationship() function can be in either table. I can specify the backref from the many class to the one class or define it as a child in the one class with the backref to the many class. Is one placement preferred over the other? If not, what criteria are used to determine where the relationship() function should be placed? I think the best use case is to put it on both, using back_populates. This is the focus of current documentation: http://docs.sqlalchemy.org/en/rel_1_0/orm/backref.html It's more verbose but I think it's clearer and functionally this is what backref does behind the scenes in any case. Rich -- 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/d/optout.
Re: [sqlalchemy] Can't locate strategy for ... (('lazy', 'joined'),)
Yes, 'number' is a column, as you surmised. When I drop that from the path it works fine. The only remaining problem is/was that this ends up loading in every field in the child_product table, and this includes a potentially massive BSON column (and more). After looking into this, I've now learned about deferred, defer, undefer, loadonly, etc. This seems to be the correct way to manage this, and it appears to be working fine: q = q.options(sa.orm .joinedload(defined_items) .joinedload(child_product) .load_only(number) ) Thanks for pointing me in the right direction! This page had the info I needed: http://docs.sqlalchemy.org/en/latest/orm/loading_columns.html Russ On Thursday, May 21, 2015 at 4:01:16 PM UTC-4, Michael Bayer wrote: On 5/21/15 3:56 PM, Russ wrote: nope. I'd need a complete, self-contained and succinct example I can run, thanks Ok, thanks. This is a beefy one so that will be extremely tricky to extract. I had hoped that the combo of lazy+joined would have been a clear indicator since they are opposite loading strategies. the word lazy there is the name of the field. It is the internal equivalent to the lazy argument on relationship: http://docs.sqlalchemy.org/en/rel_1_0/orm/relationship_api.html?highlight=relationship#sqlalchemy.orm.relationship.params.lazy that said, here is exactly what will cause your error. Any of the attribute names defined_items, child_product or number are not in fact bound to a relationship(), and instead refer to a Column-mapped attribute. Looks a lot like number here is a Column. Is that the case? You can't call joinedload for a column attribute. -- 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/d/optout.
Re: [sqlalchemy] relationsip(): In Parent or Child Class?
On Thu, 21 May 2015, Mike Bayer wrote: I think the best use case is to put it on both, using back_populates. This is the focus of current documentation: http://docs.sqlalchemy.org/en/rel_1_0/orm/backref.html It's more verbose but I think it's clearer and functionally this is what backref does behind the scenes in any case. Mike, OK. I'm carefully working through the ORM tutorial and cleaning code as I go. Thanks, Rich
[sqlalchemy] relationsip(): In Parent or Child Class?
There are a number of many-to-one table/class relationships in the application. In the many class I use ForeignKey() to relate that column to the appropriate 'one' class and column. Reading the ORM tutorial tells me that the relationship() function can be in either table. I can specify the backref from the many class to the one class or define it as a child in the one class with the backref to the many class. Is one placement preferred over the other? If not, what criteria are used to determine where the relationship() function should be placed? Rich
Re: [sqlalchemy] relationsip(): In Parent or Child Class?
On Thu, 21 May 2015, Mike Bayer wrote:
I think the best use case is to put it on both, using back_populates.
Validation check: am I correctly using relationship() in the following set
of three tables? (N.B. Other columns removed for clarity and space saving.)
class Agencies(Base):
__tablename__ = 'agencies'
name = Column(Unicode(48), Sequence('name_seq'), primary_key = True)
...
child1 = relationship('AgencyUnits', back_populates = 'agency', cascade =
'all, delete, delete-orphan')
child2 = relationship('Permits', back_populates = 'agency', cascade = 'all,
delete, delete-orphan')
child3 = relationship('Inspect', back_populates = 'agency', cascade = 'all,
delete, delete-orphan')
class AgencyUnits(Base):
__tablename__ = 'agencyUnits'
name = Column(Unicode(48), Sequence('name_seq'), primary_key = True)
parent_name = Column(Unicode(48), nullable = False, ForeignKey(\
'agencies.name'))
parent_id = relationship(Agency, back_populates = 'units')
...
child = relationship(AgencyContacts, back_populates = 'units', cascade =
'all, delete, delete-orphan')
class AgencyContacts(Base):
__tablename__ = 'agencyContacts'
contact_id = Column(Integer, Sequence('contact_id_seq'), primary_key=True)
...
unit_id = Column(Unicode(48), ForeignKey('agencyUnits.name'))
parent = relationship(AgencyUnits, back_populates = 'contacts')
Rich
[sqlalchemy] How to use make_transient() to duplicate an SQLAlchemy mapped object?
I opened a questions with example (pseudo) code on stackoverflow for that. http://stackoverflow.com/questions/30287042/how-to-use-make-transient-to-duplicate-an-sqlalchemy-mapped-object I know the question how to duplicate or copy a SQLAlchemy mapped object was asked a lot of times. The answer always depends on the needs or how duplicate or copy is interpreted. This is a specialized version of the question because I got the tip to use make_transient() for that. But I have some problems with that. I don't really know how to handle the primary key (PK) here. In my use cases the PK is always autogenerated by SQLA (or the DB in background). But this doesn't happen with a new duplicated object. -- 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/d/optout.
Re: [sqlalchemy] Can't locate strategy for ... (('lazy', 'joined'),)
On 5/21/15 3:25 PM, Russ wrote:
I have a query I am running where sqlalchemy is throwing this exception:
Exception: can't locate strategy for class
'sqlalchemy.orm.properties.ColumnProperty' (('lazy', 'joined'),)
What causes this is the addition of this joinedload_all option to a
query (q):
q = q.options(sa.orm.joinedload_all(defined_items.child_product.number))
The intent of adding that is to try and avoid emitting sql when
accessing foo.defined_items[x].child_product.number, as there are many
defined_items.
There are several other joined loads on this query that don't have an
issue. Any ideas on what would be causing this particular problem?
nope. I'd need a complete, self-contained and succinct example I can
run, thanks.
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[sqlalchemy] Can't locate strategy for ... (('lazy', 'joined'),)
I have a query I am running where sqlalchemy is throwing this exception:
Exception: can't locate strategy for class
'sqlalchemy.orm.properties.ColumnProperty' (('lazy', 'joined'),)
What causes this is the addition of this joinedload_all option to a query
(q):
q = q.options(sa.orm.joinedload_all(defined_items.child_product.number))
The intent of adding that is to try and avoid emitting sql when accessing
foo.defined_items[x].child_product.number, as there are many defined_items.
There are several other joined loads on this query that don't have an
issue. Any ideas on what would be causing this particular problem?
--
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Re: [sqlalchemy] Can't locate strategy for ... (('lazy', 'joined'),)
nope. I'd need a complete, self-contained and succinct example I can run,
thanks
Ok, thanks. This is a beefy one so that will be extremely tricky to
extract. I had hoped that the combo of lazy+joined would have been a clear
indicator since they are opposite loading strategies.
Digging through the relationship definitions I thought it might be because
the child_product relationship is explicitly declared with lazy = True
('select') and (since I don't do that on other properties) I thought
perhaps that was conflicting with the later joinedload request and
confusing the strategy lookup. However, I see that True/select is the
default, and I explicitly request joinedload at query time on other
relationships/properties without issue, so that can't be it. I'm stumped
again.
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