sqlalchemy中查询相关表

所以我有两个表“员工”和“详细信息”，如下所示。

class Employee(Base):
    __tablename__ = 'employees'
    id = Column(Integer, Sequence('employee_id_seq'), primary_key=True)
    name = Column(String(50), nullable=False)
    ............

class Detail(Base):
    __tablename__ = 'details'
    id = Column(Integer, Sequence('detail_id_seq'), primary_key=True)
    start_date = Column(String(50), nullable=False)
    email = Column(String(50))
    employee_id = Column(Integer, ForeignKey('employee.id'))
    employee = relationship("Employee", backref=backref('details', order_by=id))
    ............

现在我想做的是获取所有员工及其相应的详细信息，这是我尝试过的。

for e, d in session.query(Employee, Detail).filter(Employee.id = Detail.employee_id).all():
    print e.name, d.email

这样做的问题是它会将所有内容打印两次。我尝试使用 .join() 并打印结果两次。

我想要实现的目标是

print Employee.name
print Employee.details.email

如果您确实只关心几列，则可以直接在查询中指定它们：

q = session.query(Employee.name, Detail.email).filter(Employee.id == Detail.employee_id).all()
for e, d in q:
    print e, d

如果您确实想加载对象实例，那么我会采取不同的做法：

# query all employees
q = (session.query(Employee)
        # load Details in the same query
        .outerjoin(Employee.details)
        # let SA know that the relationship "Employee.details" is already loaded in this query so that when we access it, SA will not do another query in the database
        .options(contains_eager(Employee.details))
        ).all()

# navigate the results simply as defined in the relationship configuration
for e in q:
    print(e)
    for d in e.details:
        print(" ->", d)

至于你的duplicate结果问题，我相信你的真实代码中有一些“额外”的东西会产生这个错误......