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How to autofill table with foreign keys in SQLAlchemy using relationships?

Using SQLAlchemy I have three models: Parent1, Parent2, and Child, where Parent1 has one-to-one relationship with Parent2, and both of them has the same relationship with Child. Here are they:
from extensions import db_session
class Parent1(Base):
__tablename__ = 'parent1'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
parent2 = relationship("Parent2", backref="parent1", uselist=False)
child = relationship("Child", backref="parent1", uselist=False)
class Parent2(Base):
__tablename__ = 'parent2'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
parent1_id = Column(Integer, ForeignKey('parent1.id'))
child = relationship("Child", backref="parent2", uselist=False)
class Child(Base):
id = Column(Integer, primary_key=True)
parent1_id = Column(Integer, ForeignKey('parent1.id'))
parent2_id = Column(Integer, ForeignKey('parent2.id'))
What I am trying to achieve is to fill Child table with its parents' foreign keys.
So, when I execute this:
parent1 = Parent1(name="Adil")
parent1.parent2 = Parent2(name="Aisha")
parent1.child = Child()
db_session.add(parent1)
db_session.commit()
to the parents tables it inserts data as needed, however to the Child table it inserts data like this:
Child
id parent1_id parent2_id
1 1 NULL
How to properly set relationships, so that on any insert to the Parent1->Parent2 tables it also inserts its ids as foreign keys to Child table?
What I want to achieve is:
Child
id parent1_id parent2_id
1 1 1

I came up with this solution. I removed declared model Child and created association table between two models and added secondary parameter filling with assocation table name:
Base = declarative_base()
child = Table('child', Base.metadata,
Column('parent1_id', Integer, ForeignKey('parent1.id')),
Column('parent2_id', Integer, ForeignKey('parent2.id')))
class Parent1(Base):
__tablename__ = 'parent1'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
parent2 = relationship("Parent2", secondary='child', uselist=False)
class Parent2(Base):
__tablename__ = 'parent2'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
parent1_id = Column(Integer, ForeignKey('parent1.id'))

How can I access subclasses from upper class in sqlalchemy?

I have 3 classes;
'Company' top class its subclass 'Department' its subclass 'DepartmentalUnit'
I can access the values ​​of all classes from the 'DepartmentalUnit' class to the top class 'Company'
What I could not do and understand despite reading the document is that;;
I cannot access other subclasses from the 'company' class
class Company(Base):
__tablename__ = 'company'
id = Column(Integer, primary_key=True)
name = Column(String)
departments = relationship('Department',backref='company')
class Department(Base):
__tablename__ = 'department'
id = Column(Integer, primary_key=True)
department_name = Column(String)
company_id = Column(Integer, ForeignKey('company.id'))
departmentalunits = relationship('DepartmentalUnit', backref='department')
class DepartmentalUnit(Base):
__tablename__ = 'departmentalunit'
id = Column(Integer, primary_key=True,nullable=False)
departmental_unit_name = Column(String)
departments_id = Column(Integer, ForeignKey('department.id'))
The code from which I access the upper classes from the subclasses:
query = session.query(DepartmentalUnit)
instance = query.all()
for i in instance:
print(i.department.company.name)
print(i.department.department_name)
print(i.departmental_unit_name)
The code I can't access other subclasses from the company class:
query = session.query(Company)
instance = query.all()
for i in instance:
print(i.department.department_name)

Your last query should be used differently:
there is a typo in the name of the relationship: should be departments instead of department
given that the relationship is 1-N, the result is a list, so you should iterate over children.
This should work:
query = session.query(Company)
for company in query.all():
print(company.name)
for dep in company.departments:
print(" ", dep.department_name)
for dep_unit in dep.departmentalunits:
print(" ", dep_unit.departmental_unit_name)

I solved the problem. I added a backref to relationships and now I can access all of them from the company. Not sure if it's a correct method? However, I am currently getting the return I want. I have no unanswered request yet.
Example solved:
class Company(Base):
__tablename__ = 'company'
id = Column(Integer, primary_key=True)
name = Column(String)
departments = relationship('Department',backref='company',uselist=False)
class Department(Base):
__tablename__ = 'department'
id = Column(Integer, primary_key=True)
department_name = Column(String)
company_id = Column(Integer, ForeignKey('company.id'))
departmentalunits = relationship('DepartmentalUnit', backref='department',uselist=False)
class DepartmentalUnit(Base):
__tablename__ = 'departmentalunit'
id = Column(Integer, primary_key=True,nullable=False)
departmental_unit_name = Column(String)
departments_id = Column(Integer, ForeignKey('department.id'))
query = session.query(Company)
instance = query.all()
for i in instance:
print(f"Company: {i.name}")
print(f"Department: {i.departments.department_name}")
print(f"Department Unit: {i.departments.departmentalunits.departmental_unit_name}")
print( f"Report Category : {i.departments.departmentalunits.reportcategoryoftheunit.report_category_name}")

SQLAlchemy - querying multiple tables and returning nested objects

Suppose we have a simple one-to-many relationship between Company and Employee, is there a way to query all companies and have a list of employees in the attribute of each company?
class Company(Base):
__tablename__ = 'company'
id = Column(Integer, primary_key=True)
name = Column(String)
class Employee(Base):
__tablename__ = 'employee'
id = Column(Integer, primary_key=True)
first_name = Column(String)
last_name = Column(String)
company_id = Column(Integer, ForeignKey(Company.id))
I'm looking for something like this:
>>> result = db.session.query(Company).join(Employee).all()
>>> result[0].Employee
[<Employee object at 0x...>, <Employee object at 0x...>]
The size of result should be same as the number of rows in company table.
I tried the following and it gives tuple of objects (which makes sense) instead of nice parent / child structure:
>>> db.session.query(Company, Employee).filter(Company.id = Employee.company_id).all()
It's not hard to convert this into my desired object structure but just wanted to see if there's any shortcut.

You have to configure the relationship in the parent class:
class Company(Base):
__tablename__ = 'company'
id = Column(Integer, primary_key=True)
name = Column(String)
employees = relationship('Employee', lazy='joined') # <<< Add this line
Then you can query it without a join:
companies = session.query(Company).all()
print(companies[0].employees)
Documentation:
https://docs.sqlalchemy.org/en/13/orm/loading_relationships.html

You could do something like this:
class Company(Base):
__tablename__ = 'company'
id = Column(Integer, primary_key=True)
name = Column(String)
employees = db.session.query(Company, Employee).filter(Company.id = self.id).all()
self.employee_list = ['{0} {1}'.format(c.first_name, c.last_name) for c in employees]
Then you could access a employee name with Company.employee_list[0]

In sqlalchemy how to access anonymous field in subquery

First, I used Sqlalchemy's polymorphic architecture.
ChildA and ChildB extends Child.
ChildA has name column.
ChildB has age column.
class Parent(Base):
__tablename__ = 'parent'
id = Column(Integer, primary_key=True)
class Child(Base):
__tablename__ = 'child'
id = Column(Integer, primary_key=True)
parent_id = Column(Integer, ForeignKey('parent.id'))
parent = relationship(Parent, backref='children')
class ChildA(Child):
__tablename__ = 'child_a'
id = Column(Integer, primary_key=True)
name = Column(String(50))
class ChildB(Child):
__tablename__ = 'child_b'
id = Column(Integer, primary_key=True)
age = Column(Integer)
parent = DBSession.query(Parent).first()
subquery = parent.children.join(ChildA).subquery()
So I want to access ChildA.name column from subquery.
Something like subquery.c.ChildA.name == 'Tom'

If I understood properly what you are trying to do, you don't really need a subquery, it could be simply something like
In [13]:
f = session.query(Parent, ChildA).join(ChildA).first()
print(f.ChildA.name)
Pedro
For the use of subqueries, I would recommend you take a look to sqlalchemy tutorial.
On the other hand, I wasn't able to use the classes as you've defined them, I had to add a ForeignKey like this
class ChildA(Child):
__tablename__ = 'child_a'
id = Column(Integer, ForeignKey('child.id'), primary_key=True)
name = Column(String(50))
class ChildB(Child):
__tablename__ = 'child_b'
id = Column(Integer, ForeignKey('child.id'), primary_key=True)
age = Column(Integer)
I have no doubt that it works for you, this probably depends on the setup.
And finally, I would like to recommend you to use a column for the type of child. With this, it will be easier to recognize the children you are using. Something like this,
class Child(Base):
__tablename__ = 'child'
id = Column(Integer, primary_key=True)
parent_id = Column(Integer, ForeignKey('parent.id'))
parent = relationship(Parent, backref='children')
type = Column(String(20))
__mapper_args__ = {
'polymorphic_identity':'child',
'polymorphic_on':type
}
class ChildA(Child):
__tablename__ = 'child_a'
id = Column(Integer, ForeignKey('child.id'), primary_key=True)
name = Column(String(50))
__mapper_args__ = {
'polymorphic_identity':'child_a',
}
class ChildB(Child):
__tablename__ = 'child_b'
id = Column(Integer, ForeignKey('child.id'), primary_key=True)
age = Column(Integer)
__mapper_args__ = {
'polymorphic_identity':'child_b',
}
Please take a look to sqlalchemy docs for details.
Hope it helps.

Sqlalchemy foreign key to a subset of polymorphic classes

I am using sqlalchemy to create a structure that resembles a graph, so that there are several types of nodes and links joining them. The nodes are defined like this:
class Node(Base):
__tablename__ = 'node'
id = Column(Integer, primary_key=True)
type = Column(Unicode)
__mapper_args__ = {'polymorphic_on': type}
class InputNode(Node):
__tablename__ = 'inputnode'
id = Column(Integer, ForeignKey('node.id'), primary_key=True)
__mapper_args__ = {'polymorphic_identity': 'input'}
class ThruNode(Node):
__tablename__ = 'thrunode'
id = Column(Integer, ForeignKey('node.id'), primary_key=True)
__mapper_args__ = {'polymorphic_identity': 'thru'}
class OutputNode(Node):
__tablename__ = 'outputnode'
id = Column(Integer, ForeignKey('node.id'), primary_key=True)
__mapper_args__ = {'polymorphic_identity': 'output'}
Now I want to create a Link table which would looks something like this:
class Link(Base):
__tablename__ = 'link'
input = Column(Integer, ForeignKey('node.id', where='type IN ("input", "thru")'))
output = Column(Integer, ForeignKey('node.id', where='type IN ("thru", "output")'))
The bit I'm struggling with is how to do the where part of it, since as I've written it is not valid in sqlalchemy. I had though of using a CheckConstraint or a ForeignKeyConstraint, but I can't see how either of them could actually be used to do this.

I haven't tryed it nor am I an expert of this, but shouldn't this work?
class Link(Base):
__tablename__ = 'link'
input = Column(Integer, ForeignKey('thrunode.id'))
output = Column(Integer, ForeignKey('outputnode.id'))
First I had another idea that maybe you could have used different names for the ids and than use those, kind of like:
instead of:
class InputNode(Node):
__tablename__ = 'inputnode'
id = Column(Integer, ForeignKey('node.id'), primary_key=True)
__mapper_args__ = {'polymorphic_identity': 'input'}
this:
class ThruNode(Node):
[...]
thrunode_id = Column(Integer, ForeignKey('node.id'), primary_key=True)
[...]
and then:
class Link(Base):
__tablename__ = 'link'
input = Column(Integer, ForeignKey('node.thrunode_id'))
output = Column(Integer, ForeignKey('node.outputnode_id'))
I got the idea from here sqlalchemy docs: declarative.html#joined-table-inheritance