Hi Pedro
I do it like this
from sqlalchemy.util import KeyedTuple
visibility = KeyedTuple(['public', 'private', 'custom'], labels=['public',
'private', 'custom'])
class Test(Base):
__tablename__ = 'test'
visibility = Column(Enum(name="visibility",
*visibility._asdict().values()), nullable=False)
in your python code you can then work with this keyedtuple
if 'hello' == visibility.public:
print "yes"
else
print "no"
you see that is registered as enum type
select * from pg_type where typname = 'visibility';
select * from pg_enum where enumlabel in ('public', 'private', 'custom');
it works pretty fine
On Friday, January 24, 2014 11:24:02 AM UTC+1, Pedro Romano wrote:
>
> Executing the following test code:
>
> import sqlalchemy as sa
> from sqlalchemy.ext.declarative import declarative_base
> from sqlalchemy.dialects import postgresql
>
>
> Base = declarative_base()
>
> class MyEnum(sa.types.TypeDecorator):
> impl = postgresql.ENUM
>
> class Test(Base):
> __tablename__ = 'test'
>
> id = sa.Column(sa.Integer, primary_key=True)
> enum_column = sa.Column(MyEnum('1', '2', '3', name='my_enum_type'))
>
> engine =
> sa.create_engine('postgresql://scott:tiger@localhost:5432/sqlalchemy_test')
> Base.metadata.create_all(engine)
>
> Results in the exception (traceback omitted):
>
> sqlalchemy.exc.ProgrammingError: (ProgrammingError) type "my_enum_type"
> does not exist
> LINE 4: enum_column my_enum_type,
> ^
> '\nCREATE TABLE test (\n\tid SERIAL NOT NULL, \n\tenum_column
> my_enum_type, \n\tPRIMARY KEY (id)\n)\n\n' {}
>
> Which means the PostgreSQL enumerate type isn't being created as it would
> have been if the postgresql.ENUM had been used directly in the column
> definition instead of the decorated MyEnum.
>
> Is this behaviour by design or should the decorated column type be
> expected to also create the corresponding PostgreSQL enumerate type?
>
> Thanks in advance for any feedback regarding this.
>
> --Pedro.
>
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