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get all parent table data but only data of child which meet condition

Hi I have 2 tables like this :
parent is like this
class Policy(Base):
"""table containing details for Policies"""
__tablename__ = "UmbrellaPolicy"
id = Column(Integer, primary_key=True)
policy_id = Column(Integer, nullable=False, index=True)
user_defined_name = Column(String(255), nullable=True)
and child is like this
class Site(Base):
__tablename__ = "Site"
id = Column(Integer, primary_key=True)
policy_id = Column(Integer, ForeignKey("Policy.id"))
site_id = Column(String(32), nullable=False, index=True)
policy = relationship("Policy", backref="sites")
now I want to get all data of Policy table but data of site table only where site_id = 100 . how I will do this in orm ? I mean sql alchemy ? kindly guide
Currently doing this but it will bring all policies: who I can restrict to just get just sites which have site_id = 100
policies = (
session.query(Policy)
.join(Site)
.filter(
Policy.user_defined_name == 'yes',
)
.all()
)
will this be done using leftjoins ? but how I can use that in flask?

Try this (it's more like a pseudocode), should return id's of needed policies:
def get_policies() -> list:
return [s.policy_id for s in site.all().filter(site_id=100)]
That will return a list of policies objects

Many to many query in sqlalchemy

There are tables for my question.
class TemplateExtra(ExtraBase, InsertMixin, TimestampMixin):
__tablename__ = 'template_extra'
id = Column(Integer, primary_key=True, autoincrement=False)
name = Column(Text, nullable=False)
roles = relationship(
'RecipientRoleExtra',
secondary='template_to_role',
)
class RecipientRoleExtra(
ExtraBase, InsertMixin, TimestampMixin,
SelectMixin, UpdateMixin,
):
__tablename__ = 'recipient_role'
id = Column(Integer, primary_key=True, autoincrement=True)
name = Column(Text, nullable=False)
description = Column(Text, nullable=False)
class TemplateToRecipientRoleExtra(ExtraBase, InsertMixin, TimestampMixin):
__tablename__ = 'template_to_role'
id = Column(Integer, primary_key=True, autoincrement=True)
template_id = Column(Integer, ForeignKey('template_extra.id'))
role_id = Column(Integer, ForeignKey('recipient_role.id'))
I want to select all templates with prefetched roles in two sql-queries like Django ORM does with prefetch_related. Can I do it?
This is my current attempt.
def test_custom():
# creating engine with echo=True
s = DBSession()
for t in s.query(TemplateExtra).join(RecipientRoleExtra, TemplateExtra.roles).all():
print(f'id = {t.id}')
for r in t.roles:
print(f'-- {r.name}')
But..
it generates select query for every template to select its roles. Can I make sqlalchemy to do only one query?
generated queries for roles are without join, just FROM recipient_role, template_to_role with WHERE %(param_1)s = template_to_role.template_id AND recipient_role.id = template_to_role.role_id. Is it correct?
Can u help me?

Based on this answer:
flask many to many join as done by prefetch_related from django
Maybe somthing like this:
roles = TemplateExtra.query.options(db.joinedload(TemplateExtra.roles)).all
Let me know if it worked.

Python SQLalchemy Join lookup by User

Wrapping my head around a way to get a list of Jobs associated to a User. My DB Model goes a little something like this.
class Job(db.Model):
id = db.Column(db.Integer, primary_key=True)
# Relationship Rows
actions = db.relationship('JobAction', backref='job')
class JobAction(db.Model):
id = db.Column(db.Integer, primary_key=True)
# Linked Rows
user_id = db.Column(db.Integer, db.ForeignKey('users.id'))
# Relationship Rows
user = db.relationship('User', foreign_keys=[user_id], backref='jobactions')
I need to get a list of Jobs that are associated to a User. I can use either the User already matching a logged in users details. Or the user.id.
I was looking at something like the below, but no dice. I can see it's overly optimistic a query, but can't see what's up. Potentially a missing Join.
# Get User first.
user = User.query.filter_by(id=1).first()
# Get their Jobs
jobs = Job.query.filter_by(actions.user=user).all()
Any ideas would be greatly appreciated.
Cheers,

I'm guessing you are missing a foreign key. If your database model looked like this:
class User(db.Model):
__tablename__ = 'users'
id = db.Column(db.Integer, primary_key=True)
jobactions = db.relationship("JobAction", back_populates="user")
class Job(db.Model):
__tablename__ = 'jobs'
id = db.Column(db.Integer, primary_key=True)
jobactions = db.relationship('JobAction', backref='job')
class JobAction(db.Model):
__tablename__ = 'jobactions'
id = db.Column(db.Integer, primary_key=True)
user_id = db.Column(db.Integer, db.ForeignKey('users.id'))
job_id = db.Column(db.Integer, db.ForeignKey('jobs.id'))
user = db.relationship(User, back_populates="jobactions")
job = db.relationship(Job, back_populates="jobactions")
Then you could use:
jobs = [ jobaction.job for jobaction in user.jobactions ]

"has_many :through" construct in sqlalchemy

In rails we can simply define relationships with the has_many :through syntax in order to access 2nd, 3rd .. nth degree relations.
In SQLAlchemy however, this seems to be more difficult. I'm trying to avoid going down the route of writing joins, as I find them to be anti-patterns in trying to keep a clean code base.
My tables look like following:
class Message(db.Model):
__tablename__ = 'message'
id = db.Column(db.Integer, primary_key=True)
text = db.Column(db.String())
user_id = db.Column(db.ForeignKey("user.id"))
user = db.relationship('User', backref="messages")
class User(db.Model):
__tablename__ = 'user'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String())
class Level(db.Model):
__tablename__ = 'level'
number = db.Column(db.Integer, nullable=False, primary_key=True)
name = db.Column(db.String(), nullable=False, primary_key=True)
users = db.relationship(
"User",
secondary="user_level",
backref="levels")
class UserLevel(db.Model):
__tablename__ = 'user_level'
user_id = db.Column(db.Integer, db.ForeignKey('user.id'), primary_key=True)
number = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(), primary_key=True)
__table_args__ = (
db.ForeignKeyConstraint(
['number', 'name'],
['level.number', 'level.name']
),
)
The idea is that a user can have multiple authorisation levels (e.g. a user can be at level 1, 3 and 6 at the same time). As the data I have does not contain unique sequence numbers for available levels, I had to resort to the use of composite keys to keep the data consistent with future updates.
To get all messages for a level I can currently do something like this:
users = Level.query[0].users
for user in users:
results.append(user.messages)
return results
This gives me all users on a level. But in order to get all messages for a certain level, I have to loop through these users and append them to a results list.
What I'd like to do is:
return Level.query[0].users.messages
This is more like the syntax I am used to from rails. How would one accomplish this in flask-SQLAlchemy?

How to build many-to-many relations using SQLAlchemy: a good example

I have read the SQLAlchemy documentation and tutorial about building many-to-many relation but I could not figure out how to do it properly when the association table contains more than the 2 foreign keys.
I have a table of items and every item has many details. Details can be the same on many items, so there is a many-to-many relation between items and details
I have the following:
class Item(Base):
__tablename__ = 'Item'
id = Column(Integer, primary_key=True)
name = Column(String(255))
description = Column(Text)
class Detail(Base):
__tablename__ = 'Detail'
id = Column(Integer, primary_key=True)
name = Column(String)
value = Column(String)
My association table is (It's defined before the other 2 in the code):
class ItemDetail(Base):
__tablename__ = 'ItemDetail'
id = Column(Integer, primary_key=True)
itemId = Column(Integer, ForeignKey('Item.id'))
detailId = Column(Integer, ForeignKey('Detail.id'))
endDate = Column(Date)
In the documentation, it's said that I need to use the "association object". I could not figure out how to use it properly, since it's mixed declarative with mapper forms and the examples seem not to be complete. I added the line:
details = relation(ItemDetail)
as a member of Item class and the line:
itemDetail = relation('Detail')
as a member of the association table, as described in the documentation.
when I do item = session.query(Item).first(), the item.details is not a list of Detail objects, but a list of ItemDetail objects.
How can I get details properly in Item objects, i.e., item.details should be a list of Detail objects?

From the comments I see you've found the answer. But the SQLAlchemy documentation is quite overwhelming for a 'new user' and I was struggling with the same question. So for future reference:
ItemDetail = Table('ItemDetail',
Column('id', Integer, primary_key=True),
Column('itemId', Integer, ForeignKey('Item.id')),
Column('detailId', Integer, ForeignKey('Detail.id')),
Column('endDate', Date))
class Item(Base):
__tablename__ = 'Item'
id = Column(Integer, primary_key=True)
name = Column(String(255))
description = Column(Text)
details = relationship('Detail', secondary=ItemDetail, backref='Item')
class Detail(Base):
__tablename__ = 'Detail'
id = Column(Integer, primary_key=True)
name = Column(String)
value = Column(String)
items = relationship('Item', secondary=ItemDetail, backref='Detail')

Like Miguel, I'm also using a Declarative approach for my junction table. However, I kept running into errors like
sqlalchemy.exc.ArgumentError: secondary argument <class 'main.ProjectUser'> passed to to relationship() User.projects must be a Table object or other FROM clause; can't send a mapped class directly as rows in 'secondary' are persisted independently of a class that is mapped to that same table.
With some fiddling, I was able to come up with the following. (Note my classes are different than OP's but the concept is the same.)
Example
Here's a full working example
from sqlalchemy import create_engine, Column, Integer, String, ForeignKey
from sqlalchemy.orm import declarative_base, relationship, Session
# Make the engine
engine = create_engine("sqlite+pysqlite:///:memory:", future=True, echo=False)
# Make the DeclarativeMeta
Base = declarative_base()
class User(Base):
__tablename__ = "users"
id = Column(Integer, primary_key=True)
name = Column(String)
projects = relationship('Project', secondary='project_users', back_populates='users')
class Project(Base):
__tablename__ = "projects"
id = Column(Integer, primary_key=True)
name = Column(String)
users = relationship('User', secondary='project_users', back_populates='projects')
class ProjectUser(Base):
__tablename__ = "project_users"
id = Column(Integer, primary_key=True)
notes = Column(String, nullable=True)
user_id = Column(Integer, ForeignKey('users.id'))
project_id = Column(Integer, ForeignKey('projects.id'))
# Create the tables in the database
Base.metadata.create_all(engine)
# Test it
with Session(bind=engine) as session:
# add users
usr1 = User(name="bob")
session.add(usr1)
usr2 = User(name="alice")
session.add(usr2)
session.commit()
# add projects
prj1 = Project(name="Project 1")
session.add(prj1)
prj2 = Project(name="Project 2")
session.add(prj2)
session.commit()
# map users to projects
prj1.users = [usr1, usr2]
prj2.users = [usr2]
session.commit()
with Session(bind=engine) as session:
print(session.query(User).where(User.id == 1).one().projects)
print(session.query(Project).where(Project.id == 1).one().users)
Notes
reference the table name in the secondary argument like secondary='project_users' as opposed to secondary=ProjectUser
use back_populates instead of backref
I made a detailed writeup about this here.

Previous Answer worked for me, but I used a Class base approach for the table ItemDetail. This is the Sample code:
class ItemDetail(Base):
__tablename__ = 'ItemDetail'
id = Column(Integer, primary_key=True, index=True)
itemId = Column(Integer, ForeignKey('Item.id'))
detailId = Column(Integer, ForeignKey('Detail.id'))
endDate = Column(Date)
class Item(Base):
__tablename__ = 'Item'
id = Column(Integer, primary_key=True)
name = Column(String(255))
description = Column(Text)
details = relationship('Detail', secondary=ItemDetail.__table__, backref='Item')
class Detail(Base):
__tablename__ = 'Detail'
id = Column(Integer, primary_key=True)
name = Column(String)
value = Column(String)
items = relationship('Item', secondary=ItemDetail.__table__, backref='Detail')