SQLAlchemy 与同一张表的关系

Question

我已经声明了以下模型：

from sqlalchemy import (
    Column,
    Table,
    Integer,
    Date,
    String,
    ForeignKey,
)

from sqlalchemy import create_engine

from sqlalchemy.orm import sessionmaker

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.ext.associationproxy import association_proxy

from sqlalchemy.orm import relationship, backref

engine = create_engine('sqlite:///data.sqlite')
DBSession = sessionmaker(bind=engine)

Base = declarative_base()


lineup = Table('lineups', Base.metadata,
               Column('match_id', Integer, ForeignKey('data.id')),
               Column('player_id', Integer, ForeignKey('players.id')))


class Match(Base):
    __tablename__ = 'data'

    id = Column(Integer, primary_key=True)
    date = Column(Date)
    tournament = Column(String)
    team1 = Column(String)
    team2 = Column(String)
    team1_lineup = relationship('Player', secondary=lineup)
    team2_lineup = relationship('Player', secondary=lineup)
    best_of = Column(Integer)
    maps = relationship('Map')
    score = Column(String)


class Map(Base):
    __tablename__ = 'maps'

    id = Column(Integer, primary_key=True)
    match = Column(Integer, ForeignKey('data.id'))
    name = Column(String)
    score = Column(String)


class Player(Base):
    __tablename__ = 'players'

    id = Column(Integer, primary_key=True)
    nickname = Column(String)
    team = Column(String)

我正在以这种方式创建新的Match 对象：

match = Match(...) # all kwargs except team1_lineup and team2_lineup

p1 = Player(id=1, nickname='p1', team='team')
p2 = Player(id=2, nickname='p2', team='team')
p3 = Player(id=3, nickname='p2', team='team')

match.team1_lineup.append(p1)
match.team2_lineup.append(p2)
match.team2_lineup.append(p3)

提交新对象后，我正在查询它。

>>> from hltv.models import Match, DBSession
>>> s = DBSession()
>>> m = s.query(Match).first()
>>> m.team1_lineup
[<hltv.models.Player object at 0x7f1a93009d10>, <hltv.models.Player object at 0x7f1a93009d90>, <hltv.models.Player object at 0x7f1a93009e10>]
>>> m.team2_lineup
[<hltv.models.Player object at 0x7f1a93009d10>, <hltv.models.Player object at 0x7f1a93009d90>, <hltv.models.Player object at 0x7f1a93009e10>]

问题是m.team1_lineup 和m.team2_lineup 是一样的。我该如何解决这个问题？ 另外，如何为每个阵容分配 ID（相同球员的阵容应该有相同的 ID）？

Answer 1

我已经设法解决了我的问题。我必须将 ID 添加到阵容（我已将其重命名为 Team），并通过为这些团队提供两个不同的 ID 来指定如何加入表格。 André 发布的另一种方式解决方案。

代码如下：

from sqlalchemy import (
    Column,
    Table,
    Integer,
    Date,
    String,
    ForeignKey,
)

from sqlalchemy import create_engine

from sqlalchemy.orm import sessionmaker

from sqlalchemy.ext.declarative import declarative_base

from sqlalchemy.orm import relationship, backref

engine = create_engine('sqlite:///data.sqlite')
DBSession = sessionmaker(bind=engine)

Base = declarative_base()


class Match(Base):
    __tablename__ = 'data'

    id = Column(Integer, primary_key=True)
    date = Column(Date)
    tournament = Column(String)
    best_of = Column(Integer)
    score = Column(String)
    maps = relationship('Map')
    team1_id = Column(ForeignKey('team.id'))
    team2_id = Column(ForeignKey('team.id'))

    team1 = relationship('Team', primaryjoin='Match.team1_id == Team.id')
    team2 = relationship('Team', primaryjoin='Match.team2_id == Team.id')


class Map(Base):
    __tablename__ = 'map'

    id = Column(Integer, primary_key=True)
    match = Column(Integer, ForeignKey('data.id'))
    name = Column(String)
    score = Column(String)


lineup = Table('lineup',
               Base.metadata,
               Column('player_id', Integer, ForeignKey('player.id')),
               Column('team_id', Integer, ForeignKey('team.id')))



class Player(Base):
    __tablename__ = 'player'

    id = Column(Integer, primary_key=True)
    nickname = Column(String)


class Team(Base):
    __tablename__ = 'team'

    id = Column(Integer, primary_key=True)
    name = Column(String)

    players = relationship('Player', secondary=lineup, backref='teams')

Answer 2

我对人际关系有一些建议：

• 考虑到每场比赛都有 2(2) 个球队和 2(2) 个不同的阵容，当您对两个不同的阵容使用相同的比赛 id 时，会发生此查询错误以返回阵容；
• 我不擅长运动，但我认为你可以对模型进行更好的归一化。例如：创建一个新的模型团队

我的建议是为 Many to many 更改一些模型，如下所示：

from sqlalchemy import (
    Column,
    Table,
    Integer,
    Date,
    String,
    ForeignKey,
)

from sqlalchemy import create_engine

from sqlalchemy.orm import sessionmaker

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.ext.associationproxy import association_proxy

from sqlalchemy.orm import relationship, backref

engine = create_engine('sqlite:///data.sqlite')
DBSession = sessionmaker(bind=engine)

Base = declarative_base()

lineup = Table('lineups', Base.metadata,
               Column('team_id', Integer, ForeignKey('teams.id')),
               Column('player_id', Integer, ForeignKey('players.id')))

class Team(Base):
    __tablename__ = 'teams'
    id = Column(Integer, primary_key=True,autoincrement= True)
    name = Column(String)
    team_lineup = relationship('Player',secondary=lineup)

match_team = Table('match_teams', Base.metadata,
               Column('match_id', Integer, ForeignKey('matches.id')),
               Column('team_id', Integer, ForeignKey('teams.id')))

class Match(Base):
    __tablename__ = 'matches'
    id = Column(Integer, primary_key=True,autoincrement= True)
    date = Column(Date)
    tournament = Column(String)
    team = relationship('Team',secondary=match_team)
    best_of = Column(Integer)
    maps = relationship('Map')
    score = Column(String)

class Map(Base):
    __tablename__ = 'maps'
    id = Column(Integer, primary_key=True,autoincrement= True)
    match = Column(Integer, ForeignKey('matches.id'))
    name = Column(String)
    score = Column(String)

class Player(Base):
    __tablename__ = 'players'
    id = Column(Integer, primary_key=True,autoincrement= True)
    nickname = Column(String)

创建对象...

s = DBSession()

Base.metadata.drop_all(engine) 
Base.metadata.create_all(engine) 

t1 = Team(name="t1")
t2 = Team(name="t2")

p1 = Player()
p2 = Player()
p3 = Player()

s.add(p1)
s.add(p2)
s.add(p3)

t1.team_lineup.append(p1)
t2.team_lineup.append(p2)
t2.team_lineup.append(p3)

s.add(t1)
s.add(t2)

m = Match()
m.tournament="xpto"
m.team.append(t1)
m.team.append(t2)

s.add(m)
s.commit()

然后，你可以看到阵容和你的预期一样：

>>> m = s.query(Match).first()
>>> for t in m.team:
...     print t.team_lineup
... 
[<__main__.Player object at 0x10b57c890>]
[<__main__.Player object at 0x10b57c910>, <__main__.Player object at 0x10b57c990>]

在这种情况下，每场比赛您可以有“n”支球队。我不知道在您管理的运动中是否有可能。