Question

我是SQLAlchemy的ORM新手，我以前只使用原始SQL。我有数据库表，Label，Position和DataSet如下：

以下相应的python类：

class Label(Base):
    __tablename__ = 'Label'

    id = Column(Integer, primary_key=True)
    name = Column(String, nullable=False, unique=true)


class Position(Base):
    __tablename__ = 'Position'

    id = Column(Integer, primary_key=True)
    name = Column(String, nullable=False, unique=true)


class DataSet(Base):
    __tablename__ = 'DataSet'

    id = Column(Integer, primary_key=True)
    label_id = Column(Integer, ForeignKey('Label.id'))
    position_id = Column(Integer, ForeignKey('Position.id'))
    timestamp = Column(Integer, nullable=False)

但是在我的服务中，我没有公开那些label_id和position_id。所以我创建了一个新的课程Data，以label和position作为字符串。

# Not a full class to only show my concept
class Data:
    # data dictionary will have data 
    def __init__(self, **kwargs):
        # So it doesn't have ids. Label and Position as string
        keys = {'label', 'position', 'timestamp'}
        self.data = {k: kwargs[k] for k in keys}

    # An example of inserting data.
    # skipped detail and error handling to clarify
    def insert(self):
        session = Session()
        # get id of label and position
        # remember that it returns a tuple, not a single value
        self.data['label_id'] = session.query(Label.id).\
            filter(Label.name == self.data['label']).one_or_none()
        self.data['position_id'] = session.query(Position.id).\
            filter(Position.name == self.data['position']).one_or_none()
        # add new dataset
        self.data.pop('label')
        self.data.pop('position')
        new_data = DataSet(**self.data)
        session.add(new_data)
        session.commit()

但它看起来有些难看，我认为应该有一种更简单的方法。有没有办法使用SQLAlchemy API组合这些表类？

Answer 1

您可以使用relationships和association proxies建立从DataSet到Label和Position个对象的链接：

from sqlalchemy.orm import relationship
from sqlalchemy.ext.associationproxy import association_proxy

class DataSet(Base):
    __tablename__ = 'DataSet'

    id = Column(Integer, primary_key=True)

    label_id = Column(Integer, ForeignKey('Label.id'))
    label = relationship('Label')
    label_name = association_proxy('label', 'name')

    position_id = Column(Integer, ForeignKey('Position.id'))
    position = relationship('Position')
    position_name = association_proxy('position', 'name')

    timestamp = Column(Integer, nullable=False)

在此之后，您可以通过新属性访问与Label（及其名称）相关联的Position和DataSet个对象：

>>> d = session.query(DataSet).first()
>>> d.position
<Position object at 0x7f3021a9ed30>
>>> d.position_name
'position1'

不幸的是，插入DataSet个对象并不是那么美好。您可以为creator指定association_proxy函数，该函数可以获取名称并创建或检索相应的对象（在this answer中找到）：

def _label_creator(name):
    session = Session()
    label = session.query(Label).filter_by(name=name).first()
    if not label:
        label = Label(name=name)
        session.add(label)
        session.commit()
    session.close()
    return label

label_name = association_proxy('label', 'name', creator=_label_creator)

为两个代理指定creator个函数后，您可以通过以下方式创建新的DataSet个对象：

dataset = DataSet(
    label_name='label1',
    position_name='position2',
    timestamp=datetime.datetime.now()
)

如何在SQLAlchemy中组合两个模型类

1 个答案: