具有两个以上外键的 SQLAlchemy 关联表

Question

我有 4 个表，Person、Video、Alias，以及一个视频和人之间的关联表 video_person_association_table。

每个人都有别名。例如“Clark Kent”是一个人。他的一个别名是“超人”，另一个是“Kal-El”

A 人和他们的别名之间存在一对多关系，如下所示：

class Person(Base):
    #... other class stuff

    aliases: ["Alias"] = relationship("Alias", back_populates="person")


class Alias(Base):
    #... other class stuff

    person_id: int = Column(Integer, ForeignKey('person.id'))
    person = relationship("Person", back_populates="aliases")

另外，一个人可以有很多视频，一个视频可以有很多人，所以视频和人的关系是多对多的：

class Person(Base):
    #... other class stuff

   videos: ["Video"] = relationship("Video", secondary=video_person_association_table, back_populates="people")

class Video(Base):
    #... other class stuff
    people: list["Person"] = relationship("Person", secondary=video_person_association_table, back_populates="videos")

video_person_association_table = Table("video_person_association_table",
                                       Base.metadata,
                                       Column('video_id', ForeignKey('video.id'), primary_key=True),
                                       Column('person_id', ForeignKey('person.id'), primary_key=True)
                                       )

这很好用。但是，我想在关联表中添加第三列，如下所示：

video_person_association_table = Table("video_person_association_table",
                                       Base.metadata,
                                       Column('video_id', ForeignKey('video.id'), primary_key=True),
                                       Column('person_id', ForeignKey('person.id'), primary_key=True),
                                       Column('alias_id', ForeignKey('alias.id'), primary_key=True)
                                       )

我想这样做是因为我将人们与视频联系起来的方式是通过他们的别名。

例如，名为“这是一只鸟，这是一架飞机，不！是超人”的视频将与别名“超人”相关联。另一个视频“Kal-El the last son of Krypton”将关联 使用别名“Kal-El”

如果 Aliases 表如下所示：

id	person_id	name
1	1	Clark Kent
2	1	Superman
3	1	Kal-El

People 表如下所示： |编号 |生日| |:----:|:------:| | 1 | 1938|

视频表如下所示：

id	title
1	It's a bird, it's a plane, No! It's Superman
2	Kal-El the last son of Krypton
3	Clark Kent receives an award for excellence in journalism

我希望关联表如下所示：

video_id	person_id	alias_id
1	1	1
2	1	2
3	1	3

我想要这样的原因是，我想保留某个视频为何与某个人相关联的信息。所以为此我想存储我用来做关联的别名的id。

我还需要那里的人员 ID 以使查询更简单。如果人员 ID 不存在并且我想查看哪些人与某个视频相关联，我需要将视频加入别名，然后将别名加入人员。

这个问题是当我尝试运行 SqlAlchemy 并在这个表中插入一些东西时，我得到：

sqlalchemy.exc.IntegrityError: (sqlite3.IntegrityError) NOT NULL constraint failed: video_person_association_table.alias_id
[SQL: INSERT INTO video_person_association_table (video_id, person_id) VALUES (?, ?)]
[parameters: ((1, 1), (1, 2))]

这里的问题不在于 NOT NULL 约束。它只是尝试将两个值插入关联表。有没有办法配置关系，使其使用 3 列而不是 2 列？

我知道有一个Association Object 但它适用于更复杂的场景，您需要在关联表上存储除外键之外的额外信息。但我只需要外键。

Answer 1

我最终做的是使用Association Object 和Association Proxy，主要使用this SO 问题。

关联对象：

class VideoPersonAssociation(Base):
    """Represents an association between a Person and a video"""
    __tablename__ = 'video_person'

    id = Column(Integer, primary_key=True)
    video_id = Column(Integer, ForeignKey('video.id'), nullable=False)
    person_id = Column(Integer, ForeignKey('person.id'), nullable=False)
    alias_id = Column(Integer, ForeignKey('alias.id'), nullable=False)

    __table_args__ = (UniqueConstraint(video_id, person_id, alias_id),)

    video = relationship("Video", back_populates="people_association_objects")
    person = relationship("Person", back_populates="videos_association_objects")
    alias = relationship("Alias")

视频类：

class Video(Base):
#... other class variables

    #relationship that ties to VideoTagAssociation *video_id* foreign key. This is filled with <VideoTagAssociation objects>
    people_association_objects: list["Person"] = relationship("VideoPersonAssociation", back_populates="video")

    #proxies the people_association_objects relationship to the *person* field of VideoTagAssociation. (Allows us to do `Video.people` instead of `Video.people_association_objects.person`)
    people = association_proxy('people_association_objects', 'person')

    #relationship that ties to VideoTagAssociation *alias_id* foreign key. This is filled with <VideoTagAssociation objects> too
    #viewonly=True means that we do not intend to set aliases through this field
    people_aliases_association_objects: list["Alias"] = relationship("VideoPersonAssociation", viewonly=True)

    #same as the people proxy. Allows us to do `for alias in video.people_aliases: print(alias)` instead of 'for x in video.people_aliases_association_objects: print(x.alias)'
    people_aliases = association_proxy('people_aliases_association_objects', 'alias')

人物类：

class Person(Base):
#... other class variables

   #works the same way as above... We don't care about aliases id in Person. Only in Video. This why we don't have alias fields here.

    videos_association_objects: list["Video"] = relationship("VideoPersonAssociation", back_populates="person")

    videos = association_proxy('videos_association_objects', 'video')

这很好用，但您需要手动在关联对象中输入信息，如下所示：

clark = Person(birth_year=1937)

alias_clark_kent = Alias(name="Clark Kent")
alias_superman = Alias(name="Superman")
alias_Kal_El = Alias(name="Kal El")

clark.aliases.append(alias_clark_kent)
clark.aliases.append(alias_superman)
clark.aliases.append(alias_Kal_El)

video = Video(title="It's a bird, it's a plane, it's Superman")

session.add(video)
session.add(clark)
session.commit()

association = VideoPersonAssociation(video=video,person=clark,alias=alias_superman)

session.add(association)
session.commit()

#now we can do:

loaded_video = session.query(Video).filter(Video.title="It's a bird, it's a plane, it's Superman").first()


print(loaded_video.people[0].birth_year)

# will print 1937

print(loaded_video.people_aliases[0].name)

# will print "Superman"


#also we can do this
results = session.query(Video).filter(Video.people_aliases.any(Alias.name.like("%Super%"))).all()

#results will be all the videos that have a person alias like "Super"

结论：

我不确定这种方法是否比@Peter Mølgaard Pallesen 建议的方法更好。 （只需制作一个额外的关联表和一个额外的连接）。我已经包含了这个答案，因为它完成了我一开始提出的要求，但我不确定是否会更进一步，我会使用它而不是 Peter 建议的。