you want to use query.join() here, not joinedload.
http://www.sqlalchemy.org/trac/wiki/FAQ#ImusinglazyFalsetocreateaJOINOUTERJOINandSQLAlchemyisnotconstructingthequerywhenItrytoaddaWHEREORDERBYLIMITetc.whichreliesupontheOUTERJOIN
I need to update the links in that FAQ entry.
On Jan 18, 2011, at 12:12 PM, Alvaro Reinoso wrote:
> Hi all,
>
> It doesn't do the second filter with those queries:
>
> session.query(User).options(joinedload("channels")).filter(User.id ==
> int(userId)).filter(Channel.title !=
> "zeptextstuff.txt").order_by(Channel.titleView).first()
>
> or
>
> session.query(User).join(User.channels).filter(User.id ==
> int(userId)).filter(Channel.title !=
> "zeptextstuff.txt").order_by(Channel.titleView).first()
>
> or
>
> session.query(User).options(joinedload("channels")).filter(and_(User.id
> == int(userId), Channel.title !=
> "zeptextstuff.txt")).order_by(Channel.titleView).first()
>
> I get a user with a list of sorted channels, but I also get
> "zeptextstuff.txt" channel.
>
> Any clue??
>
> Thanks in advance!
>
> --
> 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.
>
--
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.
