Dee*_*Dee 11 sqlalchemy flask-sqlalchemy
假设您有以下简化示例模式,它使用SQLAlchemy加入表多态继承.Engineer和Analyst模特有Role关系.该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',
}
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如果我想找到Employees一个Role name在名称中有"石油"的地方,我会怎么做呢?
我尝试了很多很多方法.我最接近的是这个,它只返回Analyst匹配:
employee_role_join = with_polymorphic(EmployeeBase,
[Engineer, Analyst])
results = db.session.query(employee_role_join).join(Role).filter(Role.name.ilike('%petroleum%'))
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如果我尝试做这样的事情,我得到一个AttributeError,因为我正在搜索连接Role表的属性:
employee_role_join = with_polymorphic(EmployeeBase,
[Engineer, Analyst])
results = db.session.query(employee_role_join).filter(or_(
Engineer.role.name.ilike('%petroleum%'),
Analyst.role.name.ilike('%petroleum%')))
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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仅在列上联接analyst.role_id:
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(?)
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employee_role_join是一个AliasedClass同时公开Analyst和 的Engineer,然后我们可以使用它来创建一个 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%'))
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这会将生成的 SQL 更改为JOIN role ON engineer.role_id = role.id OR analyst.role_id = role.id
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