I can't describe the column "invited_name" (column_property). I don't know how to do this correctly.
class Worker(declarative_base()):
__tablename__ = "staff_worker_info"
id = Column(Integer, primary_key=True)
first_name = Column(String(40), nullable=False)
last_name = Column(String(40), nullable=False)
invited_id = Column(Integer, ForeignKey('staff_worker_info.id'))
invited = relationship("Worker", uselist=False, remote_side=[id], join_depth=1)
# I don't know how to describe this column
invited_name = column_property(
select([Worker.first_name]). \
where(Worker.id == invited_id).\
label('invited_n'))
I understand why this doesn't work, but I don't know how to write it differently.
I should get such a SQL query.
SELECT staff_worker_info.id, staff_worker_info.first_name staff_worker_info.last_name, staff_worker_info.invited_id,
(SELECT worker_invited.first_name
FROM staff_worker_info AS worker_invited
WHERE staff_worker_info.invited_id = worker_invited.id) AS invited_n,
FROM staff_worker_info
Might be a bit late, but I recently faced a similar question. I think your problem is quite easy to solve with only the relationship. If you want you can also solve it by using a column_property.
First, using the relationship. If you make the invited relationship joined, then the actual query that is send to the database is a self-join. You can access the first name via that relationship (reference https://docs.sqlalchemy.org/en/14/orm/self_referential.html).
class Worker(declarative_base()):
__tablename__ = "staff_worker_info"
id = Column(Integer, primary_key=True)
first_name = Column(String(40), nullable=False)
last_name = Column(String(40), nullable=False)
invited_id = Column(Integer, ForeignKey('staff_worker_info.id'))
invited = relationship("Worker", uselist=False, remote_side=[id], join_depth=1, lazy='joined')
#property
def invited_name(self):
return self.invited.first_name
Then, if the query you want to do is more complex, and it requires you to create a column_property, you can also do it as follows (reference https://docs.sqlalchemy.org/en/14/orm/mapped_sql_expr.html):
from sqlalchemy import inspect
from sqlalchemy.orm import aliased
class Worker(declarative_base()):
__tablename__ = "staff_worker_info"
id = Column(Integer, primary_key=True)
first_name = Column(String(40), nullable=False)
last_name = Column(String(40), nullable=False)
invited_id = Column(Integer, ForeignKey('staff_worker_info.id'))
invited = relationship("Worker", uselist=False, remote_side=[id], join_depth=1)
# Requires alias
worker = aliased(Worker)
inspect(Worker).add_property(
"invited_name",
column_property(
select([worker.first_name]). \
where(worker.id == Worker.invited_id)
)
)
I found a method. But he did not like it.
invited_name = column_property(
select([text("invited_table.first_name")]).
where(text("invited_table.id = staff_worker_info.invited_id")).
select_from(text("staff_worker_info AS invited_table")).
label('invited_n'))
Related
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.
I'm trying to do something I'm fairly sure is simple using AIOPG, which can use sqlalchemy's core API. My SQL is not great, so that's where I'm falling down here.
models
class School(Base):
__tablename__ = 'schools'
id = Column(Integer, primary_key=True, nullable=False)
sa_school = School.__table__
class SubjectCategory(Base):
__tablename__ = 'subject_categories'
id = Column(Integer, primary_key=True, nullable=False)
name = Column(String(63))
sa_subject_category = SubjectCategory.__table__
class Subject(Base):
__tablename__ == 'subjects'
id = Column(Integer, primary_key=True, nullable=False)
name = Column(String(63))
category = Column(Integer, ForeignKey('subject_categories.id'), nullable=False)
sa_subject = Subject.__table__
class SchoolSubject(Base):
__tablename__ = 'school_subjects'
id = Column(Integer, primary_key=True, nullable=False)
school = Column(Integer, ForeignKey('schools.id'), nullable=False)
subject = Column(Integer, ForeignKey('subjects.id'), nullable=False)
sa_school_subject = SchoolSubject.__table__
So I'm just trying to get all schools which teach subjects that have a certain subject_category ID.
Currently I have:
from sqlalchemy import select, join
school_c = sa_school.c
school_subj_c = sa_school_subject.c
async def get_schools(subject=None, subj_cat=None)
query = select(
[school_c.id, school_c.name]
).select_from(sa_school.join(sa_school_subject)
if subj_cat:
# Then I need to filter where a school_subj.subject.category == subj_cat
pass
elif subject:
query = query.where(sa_school_subject.c.subject == subj)
cur = await conn.execute(query)
return [dict(b) async for b in cur]
After a bit of faffing, I can simply do:
query = select(
[school_c.id, school_c.name]
).select_from(sa_school.join(sa_school_subject.join(sa_subject)))
Let's assume we have the following code in some Models.py file:
class Person(db.Model):
__tablename__ = 'Persons'
ID = db.Column(db.Integer, primary_key=True, nullable=False)
Name = db.Column(db.String(255), nullable=False)
class House(db.Model):
__tablename__ = 'Houses'
ID = db.Column(db.Integer,primary_key=True,nullable=False)
OwnerID = db.Column(db.Integer, nullable=False)
TenantID = db.Column(db.Integer, nullable=False)
__table_args__ = (
db.ForeignKeyConstraint(
['OwnerID'],
['Persons.ID'],
),
db.ForeignKeyConstraint(
['TenantID'],
['Persons.ID'],
),
)
OwnerBackref = db.relationship('Person', backref='OwnerBackref', lazy=True, foreign_keys=[OwnerID])
TenantBackref = db.relationship('Person', backref='TenantBackref', lazy=True, foreign_keys=[TenantID])
And we want to reflect these models using the automap base, so we have this code in another module called Database.py:
Base = automap_base()
engine = create_engine(DB_CONNECTION, pool_size=10, max_overflow=20)
db_session = scoped_session(sessionmaker(autocommit=False, autoflush=False, bind=engine))
Base.prepare(engine, reflect=True)
Person = Base.classes.Persons
House = Base.classes.Houses
Now, when I import House in some other module I want to be able to do this:
h = db_session.query(House).first()
print(h.OwnerBackref.Name)
print(h.TenantBackref.Name)
But instead I get an error saying that those 2 backrefs do not exist and instead a field called 'persons' gets added to my House object but the problem here is that it links only 1 (either the Tenant either the Owner). By this I mean that if I do this:
print(h.persons.Name)
It will only print the Name either for the respective Tenant either for the Owner leaving me with no way of accessing the informations for the other one. (Note here that the names that I set to the backrefs are nowhere to be found)
So, my question is how can I use the backrefs I created to access my desired informations ? Am I doing something wrong here ?
The error in your code is that you are using foreign_keys= to define the relationship between the tables but you are passing the local key name not the foreign key name to the function. For your code you cannot use foreign_keys= to define the relationship within the House model as there is only one possible foreign key Person.ID but two possible local keys House.OwnerID and House.TenantID. The primaryjoin= argument should be used instead to specify this.
class Person(db.Model):
__tablename__ = 'Persons'
ID = db.Column(db.Integer, primary_key=True)
Name = db.Column(db.String(255), nullable=False)
class House(db.Model):
__tablename__ = 'Houses'
ID = db.Column(db.Integer,primary_key=True)
OwnerID = db.Column(db.Integer, db.ForeignKey('Persons.ID'), nullable=False)
TenantID = db.Column(db.Integer, db.ForeignKey('Persons.ID'), nullable=False)
Owner = db.relationship('Person', backref='HousesOwned', primaryjoin='House.OwnerID == Person.ID')
Tenant = db.relationship('Person', backref='HousesOccupied', primaryjoin='House.TenantID == Person.ID')
If you placed the relationship statements in in the Person model rather than the House model then you could use either foreign_keys= or primaryjoin= to define the relationship. The following code will result in exactly the same relationships as in the previous code.
class Person(db.Model):
__tablename__ = 'Persons'
ID = db.Column(db.Integer, primary_key=True)
Name = db.Column(db.String(255), nullable=False)
HousesOwned = db.relationship('House', backref='Owner', foreign_keys='[House.OwnerID]')
HousesOccupied = db.relationship('House', backref='Tenant', foreign_keys='[House.TenantID]')
class House(db.Model):
__tablename__ = 'Houses'
ID = db.Column(db.Integer,primary_key=True)
OwnerID = db.Column(db.Integer, db.ForeignKey('Persons.ID'), nullable=False)
TenantID = db.Column(db.Integer, db.ForeignKey('Persons.ID'), nullable=False)
I am getting this strange error, and I'm saying strange because I made a change to an unrelated table.
I am trying to query my tDevice table which looks like this:
class TDevice(Base):
__tablename__ = 'tDevice'
ixDevice = Column(Integer, primary_key=True)
ixDeviceType = Column(Integer, ForeignKey('tDeviceType.ixDeviceType'), nullable=False)
ixSubStation = Column(Integer, ForeignKey('tSubStation.ixSubStation'), nullable=False)
ixModel = Column(Integer, ForeignKey('tModel.ixModel'), nullable=True)
ixParentDevice = Column(Integer, ForeignKey('tDevice.ixDevice'), nullable=True)
sDeviceName = Column(Unicode(255), nullable=False)#added
children = relationship('TDevice',
backref=backref('parent', remote_side=[ixDevice]))
device_type = relationship('TDeviceType',
backref=backref('devices'))
model = relationship('TModel',
backref=backref('devices'))
sub_station = relationship('TSubStation',
backref=backref('devices'))
and this is how I query it:
Device = DBSession.query(TDevice).filter(TDevice.ixDevice == device_id).one()
as soon as this line is executed, I get the error:
ArgumentError: relationship 'report_type' expects a class or a mapper argument (received: <class 'sqlalchemy.sql.schema.Table'>)
The only changes I've made is add a report_type relationship in my tReportTable
which now looks like this:
class TReport(Base):
__tablename__ = 'tReport'
ixReport = Column(Integer, primary_key=True)
ixDevice = Column(Integer, ForeignKey('tDevice.ixDevice'), nullable=False)
ixJob = Column(Integer, ForeignKey('tJob.ixJob'), nullable=False)
ixReportType = Column(Integer, ForeignKey('tReportType.ixReportType'), nullable=False) # added
report_type = relationship('tReportType',
uselist=False,
backref=backref('report'))
device = relationship('TDevice',
uselist=False,
backref=backref('report'))
job = relationship('TJob',
uselist=False,
backref=backref('report'))
I'm still new to SqlAlchemy so I can't seem to see how adding that relationship should be causing this error if I am iterating another table
not happy with myself since it's such a dumb mistake but here is my culprit:
report_type = relationship('tReportType',
uselist=False,
backref=backref('report'))
should be:
report_type = relationship('TReportType',
uselist=False,
backref=backref('report'))
capital T instead of t, I should be referencing the class, not my actual table name: 'tReportType' -> 'TReportType'
I am using SQLAlchemy here, trying to make a couple tables and link them and am having problems implementing this.
class Team(Base):
__tablename__ = "teams"
id = Column(Integer, primary_key=True)
espn_team_id = Column(Integer, unique=True, nullable=False)
games = relationship("Game", order_by="Game.date")
def __init__(self, name):
self.name = name
self.espn_team_id = espn_team_id
self.games = games
class Game(Base):
__tablename__ = "games"
id = Column(Integer, primary_key=True)
espn_game_id=Column(Integer, unique=True, nullable=False)
date = Column(Date)
h_espn_id = Column(Integer, ForeignKey('teams.espn_team_id'))
a_espn_id = Column(Integer, ForeignKey('teams.espn_team_id'))
I have this in one file which I use to create the tables. Then in another file I use the insert() function to put values into both tables. I think if I have a team with espn_team_id 360, and then I put in multiple games into the game table which have either h_espn_id=360, or a_espn_id=360, i should be able to do:
a = Table("teams", metadata, autoload=True)
a = session.query(a).filter(a.c.espn_team_id==360).first().games
and it should give me a list of all games team with ID 360 has played. But instead I get this error
AttributeError: 'NamedTuple' object has no attribute 'games'
What am I misunderstanding about SQLAlchemy or relational databases here?
Firstly, you don't have to create another Table object, as it is available as Team.__table__. Anyway, you can just query the mapped class, e.g.
query = Session.query(Team).filter(Team.espn_team_id == 360)
team360 = query.one()
games = team360.games
Refer to the documentation for methods .one(), .first(), and .all(): http://docs.sqlalchemy.org/en/latest/orm/query.html
Here is the solution I found, took way too long to understand this...
class Team(Base):
__tablename__ = "teams"
id = Column(Integer, primary_key=True)
name = Column(String)
espn_team_id = Column(Integer, unique=True, nullable=False)
h_games = relationship(
"Game",
primaryjoin="Game.h_espn_id==Team.espn_team_id",
order_by="Game.date")
a_games = relationship(
"Game",
primaryjoin="Game.a_espn_id==Team.espn_team_id",
order_by="Game.date")
#hybrid_property
def games(self):
return self.h_games+self.a_games
def __init__(self, name):
self.name = name
self.espn_team_id = espn_team_id
self.h_games = h_games
self.a_games = a_games
self.games = games