SQLAlchemy - How to make "django choices" using SQLAlchemy?

Question

In Django we can use very simple "choices" e.g.:

GENDER_CHOICES = (     ('M', 'Male'),     ('F', 'Female'), ) class Foo(models.Model):     gender = models.CharField(max_length=1, choices=GENDER_CHOICES) 

How to make something like this using SQLAlchemy?

Answer 1

Use custom types.

Example:

import sqlalchemy.types as types  class ChoiceType(types.TypeDecorator):      impl = types.String      def __init__(self, choices, **kw):         self.choices = dict(choices)         super(ChoiceType, self).__init__(**kw)      def process_bind_param(self, value, dialect):         return [k for k, v in self.choices.iteritems() if v == value][0]      def process_result_value(self, value, dialect):         return self.choices[value] 

The use of it would look like:

    class Entity(Base):         __tablename__ = "entity"         height = Column(             ChoiceType({"short": "short", "medium": "medium", "tall": "tall"}), nullable=False         ) 

If you are using Python 3, you have to change iteritems() to items().

Answer 2

I would probably go for sqlalchemy_utils