如何在 SQLAlchemy 中设置 M2M 混合计数属性?

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【中文标题】如何在 SQLAlchemy 中设置 M2M 混合计数属性?【英文标题】:How to set a M2M hybrid count property in SQLAlchemy? 【发布时间】:2022-01-22 20:26:03 【问题描述】:

我有两个由 M2M 关系绑定的表。书和作家,作家可以有很多书,书也可以有很多作家。

我想在书籍和作家上都拥有一个count 属性,以便我可以按例如出书最多的作家对它们进行排序。

# many to many association table
book_writer_association_table = Table('book_writer_association',Base.metadata,
                                      Column('book_id',ForeignKey('book.id'), primary_key=True),
                                      Column('Writer',ForeignKey('writer.id'), primary_key=True)
)

class Book(Base):
    __tablename__ = 'base'

    id = Column(Integer, primary_key=True)
    name = Column(String)
    writers = relationship(Writer,secondary=book_writer_association_table,back_populates="books")




class Writer(Base):
    __tablename__ = 'writer'

    id = Column(Integer, primary_key=True)
    name = Column(String)
    books = relationship(Book,secondery=book_writer_association_table,back_populates="writers")

    @hybrid_property
    def book_count(self):
        return len(self.books)

    @book_count.expression
    def book_count(cls):
        #what goes here?        

我尝试了各种方法,例如详细的here:

class Foo(Base):
    __tablename__ = 'foo'
    id = Column(Integer, primary_key=True)
    bar_id = Column(Integer, ForeignKey('bar.id'))
    bar = relationship('Bar')

class Bar(Base):
    __tablename__ = 'bar'
    id = Column(Integer, primary_key=True)

    @hybrid_property
    def foo_count(self):
        return object_session(self).query(Foo).filter(Foo.bar==self).count()

    @foo_count.expression
    def foo_count(cls):
        return select([func.count(Foo.id)]).where(Foo.bar_id == cls.id).label('foo_count')

但是,在这个例子中,只有两个表,我不确定如何在这里实现更复杂的连接。另一位用户建议使用column_property,但我在那里遇到了完全相同的问题。我不确定如何进一步将表添加到联接中。

【问题讨论】:

【参考方案1】:

您可以自定义从here 到 M2M 案例的想法。为此,您应该在 hybrid_property 中提及 association_table 而不是 Book 表。因此,您消除了与 Book 表的连接,并将您的案例简化为一对多关系。

我想出了这个解决方案。

from typing import List

from sqlalchemy import Column, ForeignKey, Integer, String, select, func, create_engine, Table
from sqlalchemy.ext.declarative import as_declarative
from sqlalchemy.ext.hybrid import hybrid_property
from sqlalchemy.orm import relationship, object_session, sessionmaker, Session

# Declare models
@as_declarative()
class Base:
    pass

book_writer_association_table = Table('book_writer_association',Base.metadata,
                                      Column('book_id',ForeignKey('book.id'), primary_key=True),
                                      Column('writer_id',ForeignKey('writer.id'), primary_key=True)
)

class Book(Base):
    __tablename__ = 'book'

    id = Column(Integer, primary_key=True)
    name = Column(String)
    writers = relationship("Writer", secondary=book_writer_association_table, back_populates="books")




class Writer(Base):
    __tablename__ = 'writer'

    id = Column(Integer, primary_key=True)
    name = Column(String)
    books = relationship("Book", secondary=book_writer_association_table, back_populates="writers")

    @hybrid_property
    def book_count(self):
        return object_session(self).query(book_writer_association_table).filter(book_writer_association_table.c.writer_id == self.id).count()

    @book_count.expression
    def book_count(cls):
        return select([func.count(book_writer_association_table.c.book_id)]).where(book_writer_association_table.c.writer_id == cls.id).label('book_count')

# Load DB schema
engine = create_engine('sqlite:///sqlite3.db')
Base.metadata.create_all(engine)
SessionLocal = sessionmaker(autocommit=True, bind=engine)
db: Session = SessionLocal()

# Creating test instances
b1 = Book(name="Book 1")
b2 = Book(name="Book 2")
db.add(b1)
db.add(b2)

w1 = Writer(name="Writer 1")
w2 = Writer(name="Writer 2")
db.add(w1)
db.add(w2)

b1.writers.append(w1)
b1.writers.append(w2)
b2.writers.append(w1)

query = db.query(Writer, Writer.book_count)
print(str(query)) # checking query
print()
writers: List[Writer] = query.all() # testing query
for writer, book_count in writers:
    print(f"writer.name: book_count")

结果:

> Writer 1: 2
> Writer 2: 1

我不确定如何进一步将表添加到联接中。

db.query(Writer, Writer.book_count) 这里的 SQL 看起来很干净,没有任何连接。所以,我认为你不应该对后续连接有任何问题。

> SELECT writer.id AS writer_id, writer.name AS writer_name, (SELECT count(book_writer_association.book_id) AS count_1 
> FROM book_writer_association 
> WHERE book_writer_association.writer_id = writer.id) AS book_count 
> FROM writer

编辑:如果您需要加入Book 表以提供额外的过滤,您可以这样做。这里我过滤了价格低于 150 的书

from typing import List

from sqlalchemy import Column, ForeignKey, Integer, String, select, func, create_engine, Table
from sqlalchemy.ext.declarative import as_declarative
from sqlalchemy.ext.hybrid import hybrid_property
from sqlalchemy.orm import relationship, object_session, sessionmaker, Session

# Declare models
@as_declarative()
class Base:
    pass

book_writer_association_table = Table('book_writer_association',Base.metadata,
                                      Column('book_id',ForeignKey('book.id'), primary_key=True),
                                      Column('writer_id',ForeignKey('writer.id'), primary_key=True)
)

class Book(Base):
    __tablename__ = 'book'

    id = Column(Integer, primary_key=True)
    name = Column(String)
    price = Column(Integer)
    writers = relationship("Writer", secondary=book_writer_association_table, back_populates="books")




class Writer(Base):
    __tablename__ = 'writer'

    id = Column(Integer, primary_key=True)
    name = Column(String)
    books = relationship("Book", secondary=book_writer_association_table, back_populates="writers")

    @hybrid_property
    def book_count(self):
        return (
            object_session(self)
                .query(book_writer_association_table)
                .join(Book, Book.id == book_writer_association_table.c.book_id)
                .filter(book_writer_association_table.c.writer_id == self.id)
                .filter(Book.price > 150)
                .count()
        )

    @book_count.expression
    def book_count(cls):
        # return select([func.count(book_writer_association_table.c.book_id)]).where(book_writer_association_table.c.writer_id == cls.id).label('book_count')
        #
        return (
            select([func.count(book_writer_association_table.c.book_id)])
                .join(Book, Book.id == book_writer_association_table.c.book_id)
                .where(book_writer_association_table.c.writer_id == cls.id)
                .filter(Book.price > 150)
                .label('book_count')
        )

# Load DB schema
engine = create_engine('sqlite:///sqlite3.db')
Base.metadata.create_all(engine)
SessionLocal = sessionmaker(autocommit=True, bind=engine)
db: Session = SessionLocal()

# Creating test instances
b1 = Book(name="Book 1", price=100)
b2 = Book(name="Book 2", price=200)
db.add(b1)
db.add(b2)

w1 = Writer(name="Writer 1")
w2 = Writer(name="Writer 2")
db.add(w1)
db.add(w2)

b1.writers.append(w1)
b1.writers.append(w2)
b2.writers.append(w1)

query = db.query(Writer, Writer.book_count)
print(str(query)) # checking query
print()
writers: List[Writer] = query.all() # testing query
for writer, book_count in writers:
    print(f"writer.name: book_count")

查询:

SELECT writer.id              AS writer_id,
       writer.name            AS writer_name,
       (SELECT count(book_writer_association.book_id) AS count_1
        FROM book_writer_association
                 JOIN book ON book.id = book_writer_association.book_id
        WHERE book_writer_association.writer_id = writer.id
          AND book.price > ?) AS book_count
FROM writer

【讨论】:

谢谢,这是个好主意!但是,如果有不止一个关联表,或者如果您不能只引用关联表,那该怎么办呢?例如,如果不是计算作者拥有的书籍数量,您需要计算其他必须连接到 book 表的东西。例如,如果您希望媒体资源仅统计超过 100 页的图书。 @Curtwagner1984 我已经根据你的情况调整了答案,请检查

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