假设您有以下简化的示例架构,该架构使用SQLAlchemy联接表多态继承。 Engineer
和Analyst
模型具有Role
关系。 Intern
模型没有。
class Role(db.Model):
__tablename__ = 'role'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(16), index=True)
class EmployeeBase(db.Model):
__tablename__ = 'employee_base'
id = db.Column(db.Integer, primary_key=True)
some_attr = db.Column(db.String(16))
another_attr = db.Column(db.String(16))
type = db.Column(db.String(50), index=True)
__mapper_args__ = {
'polymorphic_identity': 'employee',
'polymorphic_on': type
}
class Engineer(EmployeeBase):
__tablename__ = 'engineer'
id = db.Column(db.Integer, db.ForeignKey('employee_base.id'), primary_key=True)
role_id = db.Column(db.Integer, db.ForeignKey('role.id'), index=True)
role = db.relationship('Role', backref='engineers')
__mapper_args__ = {
'polymorphic_identity': 'engineer',
}
class Analyst(EmployeeBase):
__tablename__ = 'analyst'
id = db.Column(db.Integer, db.ForeignKey('employee_base.id'), primary_key=True)
role_id = db.Column(db.Integer, db.ForeignKey('role.id'), index=True)
role = db.relationship('Role', backref='analysts')
__mapper_args__ = {
'polymorphic_identity': 'analyst',
}
class Intern(EmployeeBase):
__tablename__ = 'intern'
id = db.Column(db.Integer, db.ForeignKey('employee_base.id'), primary_key=True)
term_ends = db.Column(db.DateTime, index=True, nullable=False)
__mapper_args__ = {
'polymorphic_identity': 'intern',
}
如果我想找到Employees
Role
的{{1}}名称中有“石油”的地方,我该怎么做?
我尝试了很多方法。我最接近的是这个,它只会返回name
个匹配项:
Analyst
如果我尝试执行类似的操作,则会得到employee_role_join = with_polymorphic(EmployeeBase,
[Engineer, Analyst])
results = db.session.query(employee_role_join).join(Role).filter(Role.name.ilike('%petroleum%'))
,因为我正在搜索联接的AttributeError
表的属性:
Role
答案 0 :(得分:3)
您可以尝试显式指定join ON子句,因为第一个查询的问题似乎是Role
仅在analyst.role_id
列上连接:
employee_role_join = with_polymorphic(EmployeeBase, [Engineer, Analyst])
results = session.query(employee_role_join).join(Role).filter(Role.name.ilike('%petroleum%'))
print(str(results))
SELECT employee_base.id AS employee_base_id,
employee_base.some_attr AS employee_base_some_attr,
employee_base.another_attr AS employee_base_another_attr,
employee_base.type AS employee_base_type,
engineer.id AS engineer_id,
engineer.role_id AS engineer_role_id,
analyst.id AS analyst_id,
analyst.role_id AS analyst_role_id
FROM employee_base
LEFT OUTER JOIN engineer ON employee_base.id = engineer.id
LEFT OUTER JOIN analyst ON employee_base.id = analyst.id
JOIN role ON role.id = analyst.role_id
WHERE lower(role.name) LIKE lower(?)
employee_role_join
是AliasedClass
,它同时公开Analyst
和Engineer
,然后我们可以使用它们来创建join-ON子句,如下所示:
results = session.query(employee_role_join)\
.join(Role, or_( \
employee_role_join.Engineer.role_id==Role.id, \
employee_role_join.Analyst.role_id==Role.id \
))\
.filter(Role.name.ilike('%petroleum%'))
将生成的SQL更改为JOIN role ON engineer.role_id = role.id OR analyst.role_id = role.id
答案 1 :(得分:2)
在role_id
上定义EmployeeBase
。即使Intern没有回到role
表的关系,在这种情况下该字段也可以为null。
我将EmployeeBase
更改为此:
class EmployeeBase(db.Model):
__tablename__ = 'employee_base'
id = db.Column(db.Integer, primary_key=True)
role_id = db.Column(db.Integer, db.ForeignKey('role.id'), index=True)
given_name = db.Column(db.String(16))
surname = db.Column(db.String(16))
type = db.Column(db.String(50), index=True)
__mapper_args__ = {
'polymorphic_identity': 'employee',
'polymorphic_on': type
}
并从所有其他员工模型中删除了role_id
列定义。
db.create_all()
petrolium_engineer = Role(name='Petrolium Engineer')
geotech_engineer = Role(name='Geotech Engineer')
analyst_petrolium = Role(name='Analyst of Petrolium')
db.session.add(petrolium_engineer)
db.session.add(geotech_engineer)
db.session.add(analyst_petrolium)
db.session.add(
Intern(given_name='Joe', surname='Blogs', term_ends=datetime.now())
)
db.session.add(
Engineer(given_name='Mark', surname='Fume', role=petrolium_engineer)
)
db.session.add(
Engineer(given_name='Steve', surname='Rocks', role=geotech_engineer)
)
db.session.add(
Analyst(given_name='Cindy', surname='Booker', role=analyst_petrolium)
)
db.session.commit()
petrolium_roles = db.session.query(EmployeeBase).join(Role).\
filter(Role.name.contains('Petrolium')).all()
for emp in petrolium_roles:
print(f'{emp.given_name} {emp.surname} is {emp.role.name}')
# Mark Fume is Petrolium Engineer
# Cindy Booker is Analyst of Petrolium