插入对象的最快方法(如果SQLAlchemy不存在)

Question

所以我对SQLAlchemy很新.

我有一个模型显示表中有大约10,000行.这是班级:

class Showing(Base):
    __tablename__   = "showings"

    id              = Column(Integer, primary_key=True)
    time            = Column(DateTime)
    link            = Column(String)
    film_id         = Column(Integer, ForeignKey('films.id'))
    cinema_id       = Column(Integer, ForeignKey('cinemas.id'))

    def __eq__(self, other):
        if self.time == other.time and self.cinema == other.cinema and self.film == other.film:
            return True
        else:
            return False

任何人都可以给我一些关于插入新表现的最快方法的指导,如果它还不存在的话.我觉得它稍微复杂一点,因为如果时间,电影和电影在表演中是独一无二的,那么表演才是唯一的.

我目前有这个代码:

def AddShowings(self, showing_times, cinema, film):
    all_showings = self.session.query(Showing).options(joinedload(Showing.cinema), joinedload(Showing.film)).all()
    for showing_time in showing_times:
        tmp_showing = Showing(time=showing_time[0], film=film, cinema=cinema, link=showing_time[1])
        if tmp_showing not in all_showings:
            self.session.add(tmp_showing)
            self.session.commit()
            all_showings.append(tmp_showing)

哪个有效,但似乎很慢.任何帮助深表感谢.

Answer 1

如果任何此类对象基于列的组合是唯一的,则需要将这些对象标记为复合主键.将primary_key=True关键字参数添加到每个列中,id完全删除列:

class Showing(Base):
    __tablename__   = "showings"

    time            = Column(DateTime, primary_key=True)
    link            = Column(String)
    film_id         = Column(Integer, ForeignKey('films.id'), primary_key=True)
    cinema_id       = Column(Integer, ForeignKey('cinemas.id'), primary_key=True)

这样,您的数据库可以更有效地处理这些行(不需要递增列),SQLAlchemy现在自动知道两个实例Showing是否相同.

我相信你可以将新的合并Showing到会话中:

def AddShowings(self, showing_times, cinema, film):
    for showing_time in showing_times:
        self.session.merge(
            Showing(time=showing_time[0], link=showing_time[1],
                    film=film, cinema=cinema)
        )