On Jul 25, 2012, at 4:29 PM, Stephan Hügel wrote:
>
> OK, I've done the following
>
> def create_signlist(sl):
> class SignList(db.Model):
>
> __tablename__ = listname.lower()
>
> id = db.Column("id", db.Integer(), primary_key=True)
> reference = db.Column(db.String(50), nullable=False, unique=True)
>
> def __init__(self, reference):
> self.reference = reference
>
> SignList.__name__ = listname
> return SignList
>
> signlists = ['lka', 'kal']
> for s in signlists:
> create_signlists(s)
>
> But this gives me a warning:
>
> SAWarning: The classname 'SignList' is already in the registry of this
> declarative base, mapped to <class 'glyph.models.lka'>
> _as_declarative(cls, classname, cls.__dict__)
that's only a warning, you can ignore it. It means if you have some class
which makes a relationship() to "SignList" using just the name, you won't get
this new class.
if you want to make a new class that has a new name from the start, use type():
def __init__(self, reference):
self.reference = reference
d = dict(
__tablename__ = listname.lower()
id = db.Column("id", db.Integer(), primary_key=True)
reference = db.Column(db.String(50), nullable=False, unique=True)
__init__ = __init__
)
my_class = type(listname, (db.Model,), d)
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