Flask-SqlAlchemy composite key one to many relationship Sql Server - sql-server-2012

I'm currently getting an error, I'm using sql server and trying to model a simple Parent with an array of Children:
sqlalchemy.exc.NoForeignKeysError: Could not determine join condition
between parent/child tables on relationship Parent.children- there are no
foreign keys linking these tables. Ensure that referencing columns
are associated with a ForeignKey or ForeignKeyConstraint, or specify a
'primaryjoin' expression.
my classes are set up simply as follows:
class Parent(db.Model):
__tablename__ = "parent"
parentId = db.Column(db.Integer, primary_key=True)
parentVersion = db.Column(db.Integer, primary_key=True)
children = db.relationship('Child', backref="parent",lazy=True)
class Child(db.Model):
__tablename__ = "child"
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(512), nullable=False)
parentId = db.Column(db.Integer, nullable=False)
parentVersion = db.Column(db.Integer, nullable=False)
ForeignKeyConstraint(['parentId', 'parentVersion'], ['parent.parentId', 'parent.parentVersion']
I've tried fiddling with declaring the relationship and foreign key in several ways but i always get an error, what is the correct way to do this?

Your forgot to add a foreign key:
parentId = db.Column(db.Integer, db.ForeignKey("parent.id))
There is a lot of documentation material regarding this topic, if there is still anything unclear to you.

You are missing adding the ForeignKeyConstraint to the table args, and you are using camel case, not snake case. And you don't need the __tablename__ with Flask-SQLAlchemy.
Try:
class Parent(db.Model):
id = db.Column(db.Integer, primary_key=True)
version = db.Column(db.Integer, primary_key=True)
children = db.relationship('Child', backref="parent", lazy=True)
class Child(db.Model):
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(512), nullable=False)
parent_id = db.Column(db.Integer, nullable=False)
parent_version = db.Column(db.Integer, nullable=False)
__table_args__ = (
db.ForeignKeyConstraint(
['parent_id', 'parent_version'],
['parent.id', 'parent.version']
),
)

Related

How to perform a data migration with Alembic and two versions of my Table?

I'm trying to refactor a database model; separate one column out of a table into another new one. I'd like to do this using existing SQLAlchemy Core models & Alembic. I'd also like to use server-side INSERT ... FROM SELECT ...-style query to migrate data (docs). By avoiding having to copy all the gazillion of rows to Python-world I hope to have maximum scalability, maximum performance and minimum downtime.
My problem is the programmatic use of SQLAlchemy running on two versions of the same table name in a single Metadata context. Should I resort to using an textual SQL instead? 😕
schema.py before:
class User(Base):
__tablename__ = "users"
id = Column(BigInteger, primary_key=True, autoincrement=False, nullable=False)
[...]
profile_picture_url = Column(String, nullable=True)
schema.py after:
class User(Base):
__tablename__ = "users"
id = Column(BigInteger, primary_key=True, autoincrement=False, nullable=False)
[...]
class UserProfileExtras(Base):
__tablename__ = "user_profile_extras"
user_id = Column(BigInteger, ForeignKey("users.id"), index=True, nullable=False)
profile_picture_url = Column(String, nullable=False)
So here's my attempt to create an Alembic upgrade script:
# Import the new/current-in-code models.
from ... import User, UserProfileExtras
# Define the previous User model in order to operate on the current/old schema.
class UserBeforeUpgrade(Base):
__tablename__ = "users"
id = Column(BigInteger, primary_key=True, autoincrement=False, nullable=False)
[...]
profile_picture_url = Column(String, nullable=True)
table_before_upgrade: Table = UserBeforeUpgrade.__table__
new_target_table = UserProfileExtras.__table__
[...]
def upgrade() -> None:
op.create_table(
"user_profile_extras",
sa.Column("user_id", sa.BigInteger(), autoincrement=False, nullable=False),
sa.Column("profile_picture_url", sa.VARCHAR(), nullable=False),
[...]
)
from_user_table = (select([table_before_upgrade.c.id, table_before_upgrade.c.profile_picture_url])
.where(table_before_upgrade.c.profile_picture_url != None))
insert_from = (
new_target_table.insert().from_select(
[new_target_table.c.user_id, new_target_table.c.profile_picture_url],
from_user_table)
)
op.execute(insert_from))
[...]
[...]
Error:
sqlalchemy.exc.InvalidRequestError: Table 'users' is already defined for this MetaData instance.
Specify 'extend_existing=True' to redefine options and columns on an existing Table object.

ORM query many to many to one, Flask sql-alchemy

I have three flask-sqlalchemy-models. Books - unique entries by admin:
class Book(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
book_name = db.Column(db.String(20), unique=True, nullable=False)
def __repr__(self):
return self.book_name
Bookscomp - entries by users, related to above:
class Bookscomp(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
book = db.Column(db.Integer, db.ForeignKey('book.id'))
Company - user, related to above:
class Company(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
name = db.Column(db.String(100), nullable=False)
books = db.relationship('Bookscomp', secondary=companybook, lazy='dynamic',
backref=db.backref('company'))
companybook = db.Table('companybook',
db.Column('companyid', db.Integer, db.ForeignKey('company.id'), primary_key=True),
db.Column('bookid', db.Integer, db.ForeignKey('bookscomp.id'), primary_key=True),
)
Problem: I am trying to get book_name from Book model, through Company + Bookscomp.
So a company has many books and each book has reference to general book info.
Tried like this:
company = OrganizatCompanyion.query.filter_by(id=comp.id).first()
books = company.books.all()
for item in books:
print(item.book.book_name)
#AttributeError: 'int' object has no attribute
print(item.book)
#Gives book id from book model, but I need name
Why I cannot get book_name in the above code snippet directly? And how would it be best to achieve this?
You haven't defined a relationship between Book and Bookcomp so when you ask for item.book-- it's getting the book = db.Column(db.Integer... value. Maybe tweak your Bookscomp model to be something like:
class Bookscomp(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
book_id = db.Column(db.Integer, db.ForeignKey('book.id'))
book = db.relationship('Book')

Alembic sqlalchemy.exc.NoReferencedColumnError: (Using Flask-sqlalchemy and Flask-Migrate)

Alembic keeps giving me this error when I try to migrate my schema even though the initial migration went without a hitch.
sqlalchemy.exc.NoReferencedColumnError: Could not initialize target column for ForeignKey 'dataset.datasetid' on table 'analysis': table 'dataset' has no column named 'datasetid'
Here is a part of my models.py class
class Dataset(db.Model):
DatasetID = db.Column(db.Integer, primary_key = True)
SampleID = db.Column(db.String(50), db.ForeignKey('sample.SampleID', onupdate="cascade",ondelete="restrict"), nullable=False)
UploadDate = db.Column(db.Date, nullable=False)
UploadID = db.Column(db.Integer,db.ForeignKey('uploaders.UploadID', onupdate="cascade",ondelete="restrict"), nullable=False)
UploadStatus = db.Column(db.String(45), nullable=False)
HPFPath = db.Column(db.String(500))
DatasetType = db.Column(db.String(45), nullable=False)
SolvedStatus = db.Column(db.String(30), nullable=False)
InputFile = db.Column(db.Text)
RunID = db.Column(db.String(45))
Notes = db.Column(db.Text)
analyses = db.relationship('Analysis',backref='dataset',lazy='dynamic')
data2Cohorts = db.relationship('Dataset2Cohort',backref='dataset',lazy='dynamic')
class Dataset2Cohort(db.Model):
__tablename__='dataset2Cohort'
DatasetID = db.Column(db.Integer, db.ForeignKey('dataset.DatasetID', onupdate="cascade",ondelete="cascade"), nullable=False, primary_key = True)
CohortID = db.Column(db.Integer, db.ForeignKey('cohort.CohortID', onupdate="cascade", ondelete="restrict"), nullable=False, primary_key = True)
class Analysis(db.Model):
AnalysisID = db.Column(db.String(100), primary_key = True)
DatasetID = db.Column(db.Integer, db.ForeignKey('dataset.DatasetID', onupdate="cascade",ondelete="cascade"), nullable=False)
PipelineVersion = db.Column(db.String(30))
ResultsDirectory = db.Column(db.Text)
ResultsBAM = db.Column(db.Text)
AssignedTo = db.Column(db.String(100), nullable=True)
analysisStatuses = db.relationship('AnalysisStatus', backref='analysis', lazy='dynamic')
Does anyone know why I keep getting that error even though I have the DatasetID column in the Dataset table?
Thank you,
Teja.
Found a solution.
This seems to be an issue with how Mysql 8.x versions refer to column names in the foreign key declaration - Mysql 8.x versions always use lowercase when a column is referenced in Foreign Key statements, which cause an incompatibility with sqlalchemy. This issue is discussed here
https://github.com/sqlalchemy/sqlalchemy/issues/4344
Solution is to just upgrade the sqlalchemy to the latest version (>=1.2.x)
Teja.

SQLAlchemy Find a user with a single query (probably join necessary)

I have three models which have relationships.
Firstly, the participant model.
class Participant(UserMixin, db.Model):
__tablename__ = 'participants'
id = db.Column(db.Integer, primary_key=True)
email = db.Column(db.String(64), index=True)
team_id = db.Column(db.Integer, db.ForeignKey('teams.id'))
# Relationships
team = db.relationship("Team", back_populates="members")
Secondly, the event model.
class Event(db.Model):
__tablename__ = 'events'
id = db.Column(db.Integer, primary_key=True)
Thirdly, the team model.
class Team(db.Model):
__tablename__ = 'teams'
id = db.Column(db.Integer, primary_key=True)
event_id = db.Column(db.Integer, db.ForeignKey('events.id'))
# Relationships
members = db.relationship('Participant', back_populates="team")
Several participants are allowed to have the same email address if their team is not connected to the same event.
I am looking for a query which checks if there is a participant with a given email address who is connected to a team, which is connected to the same event. I know the event.id and the email address in advance.
Pseudo code
def check(EVENTID, EMAIL):
if db.session.query(Team, Event, Participant). \
filter(Team.event_id == EVENTID). \
filter(Participant.team_id == Team.id). \
filter(Participant.email == EMAIL).first():
return true
I think it can be done with one single query using joins, but I couldn't figure it out. Please help!

nested sqlalchemy filter with parent and son

With the following scheme:
class User(Base):
id = Column(Integer, primary_key=True)
name = Column(String)
class Photo(Base):
id = Column(Integer, primary_key=True)
user_id = Column(Integer, ForeignKey(User.id), nullable=False)
user = relationship(User)
class Tag(Base):
id = Column(Integer, primary_key=True)
tag_name = Column(String)
tag_version = Column(Integer)
photo_id = Column(Integer, ForeignKey(Photo.id), nullable=False)
photo = relationship(Photo)
How do I create an SQLAlchemy query to get all the photos of a specific user, that don't have a specific tag and version.
As in "all the photos of the user with id "1234" that don't have a "cat" of version "2" tagged in them".
Also interesting would be "all the users who have at least one photo without a specific tag"
I'm using postgreSQL btw.
Here is a complete example that sets up relationships, creates some sample data, then performs your two queries.
Setup:
from datetime import datetime
from sqlalchemy import create_engine, Column, Integer, String, ForeignKey, not_
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker, relationship
engine = create_engine('sqlite:///', echo=True)
Session = sessionmaker(bind=engine)
session = Session()
Base = declarative_base(bind=engine)
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
class Photo(Base):
__tablename__ = 'photo'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
user_id = Column(Integer, ForeignKey(User.id), nullable=False)
user = relationship(User, backref='photos')
class Tag(Base):
__tablename__ = 'tag'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
photo_id = Column(Integer, ForeignKey(Photo.id), nullable=False)
photo = relationship(Photo, backref='tags')
Base.metadata.create_all()
session.add(User(name='davidism', photos=[
Photo(name='sun', tags=[Tag(name='bright'), Tag(name='day')]),
Photo(name='moon', tags=[Tag(name='bright'), Tag(name='night')])
]))
session.add(User(name='eran', photos=[
Photo(name='party', tags=[Tag(name='people'), Tag(name='night')]),
Photo(name='cat')
]))
session.commit()
Query all photos with no tags at all:
no_tags = session.query(Photo).outerjoin(Photo.tags).filter(not_(Photo.tags.any())).all()
print 'no tags: ', len(no_tags)
Query all photos without the tag 'night':
not_night = session.query(Photo).outerjoin(Photo.tags).filter(not_(Photo.tags.any(Tag.name == 'night'))).all()
print 'not night: ', len(not_night)
Assuming existance of backrefs Tag.photo = relationship(Photo, backref='tags') and
Photo.user = relationship(User, backref="photos") both can be done using any construct. This might not generate the most optimal SQL SELECT statement, but it is a very clean sqlalchemy.
Part-1: "all the photos of the user with id "1234" that don't have a "cat" of version "2" tagged in them"
def get_user_photos_without_tag(user_id, tag_name, tag_version):
qry = (session.query(Photo)
.filter(~Photo.tags.any(and_(
Tag.tag_name == tag_name,
Tag.tag_version == tag_version))
)
.filter(Photo.user_id == user_id)
)
return qry.all()
photos = get_user_photos_without_tag(1234, 'cat', 2)
Part-2: "all the users who have at least one photo without a specific tag"
def get_user_with_photos_without_tag(tag_name, tag_version):
qry = (session.query(User)
.filter(User.photos.any(
~Photo.tags.any(and_(
Tag.tag_name == tag_name,
Tag.tag_version == tag_version))
))
)
return qry.all()
res = get_user_with_photos_without_tag('cat', 2)