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如何将此SQL查询转换为SQLAlchemy

[英]How to convert this SQL Query to SQLAlchemy

我在将此SQL查询转换为SQLAlchemy时遇到问题

SELECT Recipe.id, Recipe.name, AVG(Rating.rating) AS ar FROM Recipe LEFT OUTER JOIN Rating ON Recipe.id = Rating.recipe_id GROUP BY Recipe.id ORDER BY Recipe.id ASC;

我目前有:

session.query(Recipe,func.avg(Rating.rating).label('average')).outerjoin(Rating).filter(Recipe.id == Rating.recipe_id).group_by(Recipe.id).all()

但这会返回与此类似的内容,即缺少没有任何评分的元素:

 id |       name       |         ar
----+------------------+--------------------
  1 | First Response   | 4.0000000000000000
 34 | First Response 2 | 3.0000000000000000

在命令行中的PostgreSQL查询返回时:

 id |       name       |         ar
----+------------------+--------------------
  1 | First Response   | 4.0000000000000000
 34 | First Response 2 | 3.0000000000000000
 35 | First Response 2 |
 36 | First Response 2 |

这些是我的模型:

class Recipe(Base):
    __tablename__ = 'recipe'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    prep_time = Column(Integer)
    difficulty = Column(Integer)
    vegeterian = Column(Boolean)
    user_id = Column(Integer, ForeignKey('user.id'))
    ratings = relationship("Rating")

    def __repr__(self):
        return "<Recipe(name='{}', prep_time='{}', difficulty='{}', vegeterian='{}', user='{}')>".format(
            self.name, self.prep_time, self.difficulty, self.vegeterian, self.user_id
        )

class Rating(Base):
    __tablename__ = 'rating'
    id = Column(Integer, primary_key=True)
    recipe_id = Column(Integer, ForeignKey('recipe.id'))
    rating = Column(Integer)

    def __repr__(self):
        return "<Rating(name='{}', rating='{}')>".format(self.recipe.name, self.rating)

任何帮助将不胜感激! 非常感谢你。

去掉

filter(Recipe.id == Rating.recipe_id)

因为外部联接已经基于模型之间的外键关系产生了正确的ON子句。 过滤器表达式将添加到WHERE子句中,并为没有评级的行导致UNKNOWN,因此将它们从结果中删除。

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