Declarative classes should default to __tablename__ = classname

[Former user]

A current workaround is something like:

class Defaults( DeclarativeMeta ):        
    def __init__( cls, classname, bases, dict_ ):
        if not hasattr( cls, "__tablename__" ):
            cls.__tablename__ = classname
        return DeclarativeMeta.__init__( cls, classname, bases, dict_ )

Base = declarative_base( metadata = metadata, metaclass = Defaults )

class Cover( Base ):
    #~ __tablename__ = 'Cover'
    id        = Column( Integer, primary_key = True )
    title     = Column( Text )

i.e. The default class makes specifying !tablename! unnecessary.

At the moment if a !tablename! is not specified, a confusing error message is given:

sqlalchemy.exc.ArgumentError: Mapper 'Mapper|Cover|None' does not have a mapped_table specified.  (Are you using the return value of table.create()?  It no longer has a return value.)

[Mike Bayer]

Take the code you've put there, and add it to UsageRecipes. SQLA doesn't do any kind of derived naming stuff (with the exception of index=True on Column, which I'm not thrilled about). For example, I'd never name a table Cover (or any kind of CamelCase name) - the case sensitive name would be quoted all the time and be painful to use...so already, our disagreement on naming preference becomes a non-issue by keeping this as a usage recipe.

It's true that we do some implicit naming in declarative when we say id = Column(Integer), but that's really just the "reverse" of the fact that SQLA names the mapped attributes the same as the existing column. We don't have a corresponding make_me_a_class(Table("Cover")) which magically places a class called Cover in the global namespace.

[Former user]

zzzeek, thanks for your detailed response, which makes perfect sense. Much appreciated.

[Mike Bayer]

Removing milestone: 0.5.xx (automated comment)