I have a piece of working code but it is very inefficient, instead of a single query with a join. I get one initial query, followed by one query per row in the response.
I have to following scenario:
class Job(Base, SerializeMixin, JobInterface):
__tablename__ = 'job_subjobs'
id = Column(Integer, primary_key=True, autoincrement=True)
group_id = Column(Integer, ForeignKey("job_groups.id"), nullable=False)
class Crash(Base, SerializeMixin):
__tablename__ = 'crashes'
id = Column(Integer, primary_key=True, autoincrement=True)
job_id = Column(Integer, ForeignKey("job_subjobs.id", ondelete='CASCADE'), nullable=False)
job = relationship('Job', backref='Crash')
#hybrid_property
def job_identifier(self):
return "{}:{}".format(self.job.group_id, self.job.id)
So given the above and I perform a query for all Crashes, It will perform one SELECT for all crashes. When I iterate and ask for job_identifier it will then do one separate SELECT for each crash.
self.session.query(Crash).all()
Is there someway i can create a #hybrid_property referencing a different table and have it JOIN from the beginning and preload the expression?
I've experimented with #xxx.expression without success. If all else fails I can add another foreign key in Crash table, but I would like to avoid changing current data structure if possible.
ended up using:
jobs = relationship('Job', backref='Crash', lazy='joined')
Related
I have two models with a simple FK relationship, Stock and Restriction (Restriction.stock_id FK to Stock).
class Restriction(Model):
__tablename__ = "restrictions"
id = db.Column(db.Integer, primary_key=True)
stock_id = FK("stocks.id", nullable=True)
name = db.Column(db.String(50), nullable=False)
class Stock(Model):
__tablename__ = "stocks"
id = db.Column(db.Integer, primary_key=True)
ticker = db.Column(db.String(50), nullable=False, index=True)
I would like to retrieve Restriction object and related Stock but only Stock's ticker (there are other fields omitted here). I can simply do this with:
from sqlalchemy.orm import *
my_query = Restriction.query.options(
joinedload(Restriction.stock).load_only(Stock.ticker)
)
r = my_query.first()
I get all columns for Restriction and only ticker for Stocks with above. I can see this in the SQL query run and also I can access r.stock.ticker and see no new queries run as it is loaded eagerly.
The problem is I cannot filter on stocks now, SQLAlchemy adds another FROM clause if I do my_query.filter(Stock.ticker == 'GME'). This means there is a cross product and it is not what I want.
On the other hand, I cannot load selected columns from relationship using join. ie.
Restriction.query.join(Restriction.stock).options(load_only(Restriction.stock))
does not seem to work with relationship property. How can I have a single filter and have it load selected columns from relationship table?
SQL I want run is:
SELECT restrictions.*, stocks.ticker
FROM restrictions LEFT OUTER JOIN stocks ON stocks.id = restrictions.stock_id
WHERE stocks.ticker = 'GME'
And I'd like to get a Restriction object back with its stock and stock's ticker. Is this possible?
joinedload basically should not be used with filter. You probably need to take contains_eager option.
from sqlalchemy.orm import *
my_query = Restriction.query.join(Restriction.stock).options(
contains_eager(Restriction.stock).load_only(Stock.ticker)
).filter(Stock.ticker == 'GME')
r = my_query.first()
Because you are joining using stock_id it will also be in the results as Stock.id beside Stock.ticker. But other fields would be omitted as you wish.
I have written short post about it recently if you are interested: https://jorzel.hashnode.dev/an-orm-can-bite-you
Imagine the following (example) datamodel:
class Organization(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
friendly_name = db.Column(db.Text, nullable=False)
users = db.relationship('Users', back_populates='organizations')
groups = db.relationship('Groups', back_populates='organizations')
class User(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
organization_id = db.Column(db.Integer, db.ForeignKey('organizations.id'))
organizations = relationship("Organization", back_populates="users")
class Group(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
organization_id = db.Column(db.Integer, db.ForeignKey('organizations.id'))
organizations = relationship("Organization", back_populates="groups")
(so basically an Organization has User and Group relationships)
What we want is to retrieve the counts for users and groups. Result should be similar to the following:
id
friendly_name
users_count
groups_count
1
o1
33
3
2
o2
12
2
3
o3
1
0
This can be achieved with a query similar to
query = db.session.query(
Organization.friendly_name,
func.count(User.id.distinct()).label('users_count'),
func.count(Group.id.distinct()).label('groups_count'),
) \
.outerjoin(User, Organization.users) \
.outerjoin(Group, Organization.groups) \
.group_by(Organization.id)
which seems quite overkill. The first intuitive approach would be something like
query = db.session.query(
Organization.friendly_name,
func.count(distinct(Organization.users)).label('users_count'),
func.count(distinct(Organization.groups).label('groups_count'),
)# with or without outerjoins
which is not working (Note: With one relationship it would work).
a) Whats the difference between User.id.distinct() and distinct(Organization.users) in this case?
b) What would be the best/most performant/recommended way in SQLAlchemy to get a count for each relationship an Object has?
Bonus): If instead of Organization.friendly_name the whole Model would be selected (...query(Organization, func....)) SQLAlchemy returns a tuple with the format t(Organization, users_count, groups_count) as result. Is there a way to just return the Organization with the two counts as additional fields? (as SQL would)
b:
You can try a window function to count users and groups with good performance:
query = db.session.query(
Organization.friendly_name,
func.count().over(partition_by=(User.id, Organization.id)).label('users_count')
func.count().over(partition_by=(Group.id, Organization.id)).label('groups_count')
)
.outerjoin(User, Organization.users)
.outerjoin(Group, Organization.groups)
bonus:
To return count as a field of Organization, you can use hybrid_property, but you would not be happy with the performance.
I try to add some contribution on 3c7's cool project, and I want to apply a filter on a join query(sqlalchemy).
Simple statement: multiple rules(table) can have multiple tags(table) - I want to filter out some rules based on some tags.
[Rules][2] table (rule_id, etc)
[Tags][2] Table (tag_id, etc)
tags_rules(Junction table) (rule_id,tag_id) -- no declaration
Issue: Applying a filter after join of course that will remove only the rules that have only one tag(the one that I specify). If a rule has multiple tags, one record from the join result will be removed, but the rule will still appear in there are any other tags associated with that rule
Sql alchemy declaration:
class Rule(Base):
id = Column(Integer, primary_key=True, index=True, autoincrement=True)
name = Column(String(255), index=True)
meta = relationship("Meta", back_populates="rule", cascade="all, delete, delete-orphan")
strings = relationship("String", back_populates="rule", cascade="all, delete, delete-orphan")
condition = Column(Text)
imports = Column(Integer)
tags = relationship("Tag", back_populates="rules", secondary=tags_rules)
ruleset_id = Column(Integer, ForeignKey("ruleset.id"))
ruleset = relationship("Ruleset", back_populates="rules")
class Tag(Base):
id = Column(Integer, primary_key=True, autoincrement=True, index=True)
name = Column(String(255), index=True)
rules = relationship("Rule", back_populates="tags", secondary=tags_rules)
I tried with sub queries but the most feasible seem to apply a filter on the tags table before the join.
Current implementation:
rules = rules.select_from(Tag).join(Rule.tags).filter(~Tag.name.in_(tags))
Any idea is greatly appreciated.
I'm using flask-sqlalchemy. My class is:
class Game(db.Model):
userid = db.Column(db.Integer, primary_key=True)
gameid = db.Column(db.Integer, primary_key=True)
gameid is set to auto-increment in the database.
When I run this function:
#app.route("/requestgame")
def requestgame():
game = Game()
game.userid = session["userid"]
db.session.add(game)
db.session.commit()
session["gameid"] = game.gameid
return "gameid {}".format(game.gameid)
I get "ObjectDeletedError: Instance '' has been deleted, or its row is otherwise not present."
How can I get the gameid to return?
It looks like someone asked the same question about regular SQLAlchemy here:
SQLAlchemy Obtain Primary Key With Autoincrement Before Commit
but I have tried the recommended solution (calling session.flush()) and it doesn't seem to make any difference.
Answer no longer needed as I changed focus in code. (see my comment in answer) Post answers for future reference...
How do I retrieve results from a one to many backref ordered by a field in the child? I need all somethings for the gid ordered by index. But at this time they are retrieved randomly even though they are ordered in the ms sql server.
I have in TurboGears 2 datamodels.py:
`class Parcel(DeclarativeBase):
__tablename__ = 'GENERAL'
__table_args__ = ({'autoload': True})
gid = Column(Integer, primary_key=True)`
somethings = relationship('Something', backref='Parcel')
'class Something(DeclarativeBase):
__tablename__ = 'SKETCH'
__table_args__ = ({'autoload': True})
gid = Column(Integer, ForeignKey('GENERAL.gid'), primary_key=True)
index = Column(Integer, primary_key=True)
In Turbogears root.py:
query = DBSession.query(Parcel)
query = query.options(joinedload('somethings')
query=session.filter(Parcel.gid==gid)
Returns all somethings for gid unordered.
DBSession.query(Something).filter_by(gid=gid).order_by(Something.index).all()
edit: relationship() accepts a keyword argument order_by to order instances when you use the relationship. If you want to specify the ordering for the reverse direction, you can use the backref() function instead of the backref keyword and use the same order_by keyword argument as with relationship().