Question

请帮我在SQLAlchemy中对此进行建模。用户可以创建问题。一个问题可以有多种选择（例如，是，不，以后，可能，不知道，下一年，不适用）。我在问题和选择之间创建了一个映射器。如何在SQLAlchemy中建模响应？

question_choices = Table('question_choices', Base.metadata,
    Column('id', Integer, primary_key=True),
    Column('question_id', Integer, ForeignKey('questions.id')),
    Column('choice_id', Integer, ForeignKey('choices.id'))
    )

class Choice(Base):
    __tablename__ = 'choices'
    id = Column(Integer, primary_key=True)
    value = Column(String(30), nullable=False)

class Question(Base):
    __tablename__ = 'questions'
    id = Column(Integer, primary_key=True)
    title   = Column(String(100))
    created = Column(DateTime)

    choices = relationship('Choice', secondary=question_choices)

class questionResponse(Base):
    """A friend's response to a question"""
    __tablename__ = 'question_responses'
    id = Column(Integer, primary_key=True)
    question_id = Column(Integer, ForeignKey('questions.id'))
    choice_id = Column(Integer, ForeignKey('choices.id'))
    user_id = Column(Integer, ForeignKey('users.id'))
    created = Column(DateTime)

questionResponse模型未规范化。 Question_id和listing_id重复出现。我在mapper表中没有关系。我希望能够计算给定问题的答案。

Answer 1

QuestionResponse的映射器已经相当不错了，但它不会限制未配置选项的答案：因此，如果不允许LATER，请回答问题Will you marry me? ，数据库不限制此。

一个解决方案就是要向QuestionResponse添加两列外键约束：

class QuestionResponse(Base):
    """A friend's response to a question"""
    __tablename__ = 'question_responses'
    id = Column(Integer, primary_key=True)
    question_id = Column(Integer, ForeignKey('questions.id'))
    choice_id = Column(Integer, ForeignKey('choices.id'))
    # ...
    __table_args__ = (
            ForeignKeyConstraint(['question_id', 'choice_id'], ['question_choices.question_id', 'question_choices.choice_id']),
            )

替代（更规范化的数据库模型）是仅将FK定义为question_choices.id：

class QuestionResponse(Base):
    """A friend's response to a question"""
    __tablename__ = 'question_responses'
    id = Column(Integer, primary_key=True)
    question_choice_id = Column(Integer, ForeignKey('question_choices.id'))

edit-1 ：在这种情况下，您可以在下面定义的Question和QuestionResponse之间建立关系，这也会为您提供计数：

class Question(Base):
    # ....
    answers = relationship("QuestionResponse", 
        primaryjoin="Question.id==question_choices.c.question_id",
        secondary=question_choices,
        secondaryjoin="question_choices.c.id==QuestionResponse.question_choice_id",
        backref="question",
        )

在任何情况下，您都可以在UniqueConstraint列的question_choices表格中添加(question_id, choice_id)。

现在，为了计算回复，您可以在Question和QuestionResponse之间添加关系并返回len(answers)，也可以在Question上创建基于查询的属性：

class Question(Base):
    # ...
    answer_count = column_property(
                select([func.count(QuestionResponse.__table__.c.id)]).
                where(question_choices.c.question_id==id).
                where(question_choices.c.id==QuestionResponse.__table__.c.question_choice_id)
            )

SqlAlchemy进行了多对多的数据建模

1 个答案: