I see... I'll work something out.
Thank you Mr. Bayer!!
2011/3/1 Michael Bayer <mike...@zzzcomputing.com>:
>
> On Mar 1, 2011, at 5:50 PM, Hector Blanco wrote:
>
> Hello everyone:
>
> Let's say I have a class "User" and a class "UserGroup". One user can
> belong to one userGroup, an a userGroup can contain several users
> (pretty typical structure). It's a simple relationship I got modeled
> like:
>
> class UserGroup(declarativeBase):
> """Represents a group of users with the same features"""
> __tablename__ = "user_groups"
>
> id = Column("id", Integer, primary_key=True)
> name = Column("name", String(50))
> users = relationship("User", order_by=lambda:User.userName,
> cascade="all, delete", collection_class=set)
>
> class User(declarativeBase):
> """Represents a user"""
> __tablename__ = "users"
>
> id = Column("id", Integer, primary_key=True)
> firstName = Column("first_name", String(50))
> lastName = Column("last_name", String(50))
> email = Column("email", String(60))
> userName = Column("user_name", String(50), unique=True,
> nullable=False)
> password = Column("password", String(64), nullable=False)
> userGroupId = Column("user_group_id", Integer,
> ForeignKey("user_groups.id"))
>
> userGroup = relationship("UserGroup", uselist=False)
>
> I am working in a tool that accepts generic queries, and, basically, I
> can do something like:
>
> session.query(User.User).filter(User.User.id > 3).values("userName")
>
> And get tuples with a .userName field with all the userNames of the
> users whose id is > 3
>
> But if I try:
> session.query(User.User).filter(User.User.id > 3).values("userGroup")
>
> well yes values() accepts only scalar columns (and also you should pass the
> attribute, not a string, guess the docs aren't crystal clear on that).
>
>
> So here's the question:
>
> Is there any way of getting the "userGroup" value somehow "starting"
> (or querying) User objects? (or what would be the best way, if there
> are many ways)
>
> typically the columns you're retrieving are the thing you're "starting"
> from:
>
> query(UserGroup).join(UserGroup.users).filter(User.id > 3).all()
> if you have a lot more join going on and really need a certain entity in the
> left, you can say:
> query(UserGroup).select_from(User).join(User.userGroup).filter(User.id >
> 3).all()
> There's a ticket somewhere to allow query() to also accept a relationship()
> attribute that is specifically many-to-one, but that's just a small
> syntactic convenience. query() in general accepts entities and column
> expressions only.
>
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