sqlalchemy 如何做 choice field 选择, like django 的choice

Question

django 下的module有这样的类型选择:

class Article(models.Model):

    STATUS_CHOICES = (
        ('d', 'Draft'),
        ('p', 'Published'),
    )    
    status = models.CharField('文章状态', max_length=1, choices=STATUS_CHOICES)
    

那么在sqlalchemy 下如何做到跟djano一样的功能.

Answer 1

同样的问题，刚刚搜到了 SQLAlchemy-Utils
ChoiceType

from sqlalchemy_utils.types.choice import ChoiceType

class User(Base):
    TYPES = [
        (u'admin', u'Admin'),
        (u'regular-user', u'Regular user')
    ]

    __tablename__ = 'user'
    id = sa.Column(sa.Integer, primary_key=True)
    name = sa.Column(sa.Unicode(255))
    type = sa.Column(ChoiceType(TYPES))