I have been fighting this error for a few days now in a variety of configurations.
I tried adding Relationship lines under each of the ForeignKeys with no luck.
I also swapped my queries from using the declarative_base style calls to SQL queries.
The error only appears once I add the second ForeignKey to Task. (I originally had several foreign keys in Task and Project but pared things down until it worked and tried slowly adding things back. Adding the second ForeignKey did it no matter which table it was added to.)
My current Flask method is:
#app.route('/project/<int:project_id>/')
def showProject(project_id):
project = session.query(Project).from_statement(text("SELECT * FROM project WHERE project.id = project_id ORDER BY project.projnum DESC"))
tasks = session.query(Task).filter_by(project_id = project_id).all()
return render_template('project.html', project = project, tasks = tasks)
And I get a "no such column" error:
OperationalError: (sqlite3.OperationalError) no such column: task.project_id [SQL: u'SELECT task.id AS task_id, task.name AS task_name, task.description AS task_description, task.assigned_id AS task_assigned_id, task.project_id AS task_project_id, task.due AS task_due \nFROM task \nWHERE task.project_id = ?'] [parameters: (2,)]
This is the setup file for my database:
import os
import sys
import time
from sqlalchemy import Column, ForeignKey, Integer, String, Float
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship
from sqlalchemy import create_engine
Base = declarative_base()
engine = create_engine('sqlite:///gmnpm.db')
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
name = Column(String(250), nullable=False)
email = Column(String(250), nullable=False)
picture = Column(String(250))
class Task(Base):
__tablename__ = 'task'
id = Column(Integer, primary_key=True)
name = Column(String(250), nullable=False)
description = Column(String(2000))
assigned_id = Column(Integer, ForeignKey('user.id'))
project_id = Column(Integer, ForeignKey('project.id'))
due = Column(Integer, default=int(time.time()))
class Project(Base):
__tablename__ = 'project'
id = Column(Integer, primary_key=True)
name = Column(String(250), nullable=False)
description = Column(String(1000))
projnum = Column(Integer)
Base.metadata.create_all(engine)
After reading a bunch of documentation and going crazy trying random things on the off chance they work, I'm grateful for any suggestions!
Related
We use SQLAlchemy to read/write data, but not create tables (as it done by DBAs). Due to this, some of the definitions which are incorrect have not yet been caught (though it works for reads/writes).
Is there a way to override them for testing purposes (create tables on the fly etc.) without touching the original definition? A simple class-overriding doesn't seem to work, and I don't see any other solution to this problem:
from sqlalchemy import Column, Integer, Numeric, String
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class Product(Base):
__tablename__ = "product"
idn = Column(Numeric, primary_key=True) # should be Integer
code = Column(String, primary_key=True) # should be unique (not primary)
class Product(Product):
__tablename__ = "product"
__table_args__ = {"extend_existing": True} # want to override, not extend
idn = Column(Integer, primary_key=True)
code = Column(String, unique=True)
How can i link both entities with relationship with flask python?
for example i have this entity, here i am trying to link with user = relationship('User'), so i am getting error relation of relationship (btw: Grant, User, Client are in differents files )
from sqlalchemy.orm import relationship
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
class Grant(db.Model):
id = db.Column(db.Integer, primary_key=True)
user_id = db.Column(
db.Integer, db.ForeignKey('user.id', ondelete='CASCADE')
)
user = relationship('User')
this is the error:
sqlalchemy.exc.InvalidRequestError: When initializing mapper mapped class Grant->grant, expression 'User' failed to locate a name ('User'). If this is a class name, consider adding this relationship() to the <class 'model.Grant.Grant'> class after both dependent classes have been defined.
note: those are my anothers entities:
class User(db.Model):
id = db.Column(db.Integer, primary_key=True)
username = db.Column(db.String(40), unique=True, index=True,
nullable=False)
def check_password(self, password):
return True
and this is the Client.py
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
class Client(db.Model):
name = db.Column(db.String(40))
client_id = db.Column(db.String(40), primary_key=True)
this is the error user = relationship('User') please helpme
I'm using a bidirectional association_proxy to associate properties Group.members and User.groups. I'm having issues with removing a member from Group.members. In particular, Group.members.remove will successfully remove an entry from Group.members, but will leave a None in place of the corresponding entry in User.groups.
More concretely, the following (minimal-ish) representative code snippet fails its last assertion:
import sqlalchemy as sa
from sqlalchemy.orm import Session
from sqlalchemy.ext.associationproxy import association_proxy
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class Group(Base):
__tablename__ = 'group'
id = sa.Column(sa.Integer, autoincrement=True, primary_key=True)
name = sa.Column(sa.UnicodeText())
members = association_proxy('group_memberships', 'user',
creator=lambda user: GroupMembership(user=user))
class User(Base):
__tablename__ = 'user'
id = sa.Column(sa.Integer, autoincrement=True, primary_key=True)
username = sa.Column(sa.UnicodeText())
groups = association_proxy('group_memberships', 'group',
creator=lambda group: GroupMembership(group=group))
class GroupMembership(Base):
__tablename__ = 'user_group'
user_id = sa.Column(sa.Integer, sa.ForeignKey('user.id'), primary_key=True)
group_id = sa.Column(sa.Integer, sa.ForeignKey('group.id'), primary_key=True)
user = sa.orm.relationship(
'User',
backref=sa.orm.backref('group_memberships', cascade="all, delete-orphan"))
group = sa.orm.relationship(
'Group',
backref=sa.orm.backref('group_memberships', cascade="all, delete-orphan"),
order_by='Group.name')
if __name__ == '__main__':
engine = sa.create_engine('sqlite://')
Base.metadata.create_all(engine)
session = Session(engine)
group = Group(name='group name')
user = User(username='user name')
group.members.append(user)
session.add(group)
session.add(user)
session.flush()
assert group.members == [user]
assert user.groups == [group]
group.members.remove(user)
session.flush()
assert group.members == []
assert user.groups == [] # This assertion fails, user.groups is [None]
I've tried to follow the answers to SQLAlchemy relationship with association_proxy problems and How can SQLAlchemy association_proxy be used bi-directionally? but they do not seem to help.
I discovered your problem almost entirely by accident, as I was trying to figure out what's going on.
Because there wasn't any data in the db, I added a session.commit(). It turns out that (from the linked answer):
The changes aren't persisted permanently to disk, or visible to other transactions until the database receives a COMMIT for the current transaction (which is what session.commit() does).
Because you are just .flush()ing the changes, sqlalchemy never re-queries the database. You can verify this by adding:
import logging
logging.getLogger('sqlalchemy').setLevel(logging.INFO)
logging.getLogger('sqlalchemy').addHandler(logging.StreamHandler())
And then simply running your code. It will display all of the queries that are run as they happen. Then you can change session.flush() to session.commit() and then re-run, and you'll see that several SELECT statements are run after your commit.
It looks like either session.expire(user) or session.refresh(user) will force a refresh of the user, as well. I'm not sure if there's a way to force the update to propagate to the other object without being explicit about it (or if that's even desirable).
Trying to create an association relationship between the classes Note and Document. The problem I'm facing is that my secondary relationship only works when I use the association table object and not that table name. What I mean is that the relationship:
notes = relationship(u'Note', secondary=t_Documented, backref='documents')
works but the following does NOT work:
notes = relationship(u'Note', secondary='Documented', backref='documents')
When querying, I get the error:
sqlalchemy.exc.InvalidRequestError: When initializing mapper
Mapper|Document|Document, expression 'Documented' failed to locate a
name ("name 'Documented' is not defined"). If this is a class name,
consider adding this relationship() to the
class after both dependent classes have been defined.
I would rather use the name as my model is generated using sqlacodegen.
Moreover, SQLAlchemy docs say I can use the name (http://docs.sqlalchemy.org/en/rel_0_9/orm/relationships.html#many-to-many) with caveta "with the declarative extension in use". I Googled the term which led me to this. Question is how in my case can I augment the Base.
# model.py
# coding: utf-8
from sqlalchemy import Column, Date, DateTime, ForeignKey, ForeignKeyConstraint, Index, Integer, Numeric, String, Table, Text, text
from sqlalchemy.orm import backref, relationship
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
metadata = Base.metadata
t_Documented = Table(
'Documented', metadata,
Column('id', Integer, primary_key=True),
Column('note_id', ForeignKey(u'MySchema.Note.id'), nullable=False),
Column('document_id', ForeignKey(u'MySchema.Document.id'), nullable=False, index=True),
Column('inserted', DateTime, nullable=False, server_default=text("'0000-00-00 00:00:00'")),
Column('updated', DateTime, nullable=False, server_default=text("'0000-00-00 00:00:00'")),
Index('Documented_AK1', 'note_id', 'document_id'),
schema='MySchema'
)
class Note(Base):
__tablename__ = 'Note'
__table_args__ = {u'schema': 'MySchema'}
id = Column(Integer, primary_key=True)
class Document(Note):
__tablename__ = 'Document'
__table_args__ = {u'schema': 'MySchema'}
id = Column(ForeignKey(u'MySchema.Note.id'), primary_key=True)
title = Column(String(100), nullable=False)
author = Column(String(100), nullable=False)
notes = relationship(u'Note', secondary='Documented', backref='documents')
Using SQLAlchemy 0.9.4 and Python 2.6.6. Connector is MySQLDB and I'm using MySQL database.
your table has a schema of "MySchema" so that has to be part of it:
notes = relationship(u'Note', secondary='MySchema.Documented', backref='documents')
Am trying to setup a postgresql table that has two foreign keys that point to the same primary key in another table.
When I run the script I get the error
sqlalchemy.exc.AmbiguousForeignKeysError: Could not determine join condition between parent/child tables on relationship Company.stakeholder - there are multiple foreign key paths linking the tables. Specify the 'foreign_keys' argument, providing a list of those columns which should be counted as containing a foreign key reference to the parent table.
That is the exact error in the SQLAlchemy Documentation yet when I replicate what they have offered as a solution the error doesn't go away. What could I be doing wrong?
#The business case here is that a company can be a stakeholder in another company.
class Company(Base):
__tablename__ = 'company'
id = Column(Integer, primary_key=True)
name = Column(String(50), nullable=False)
class Stakeholder(Base):
__tablename__ = 'stakeholder'
id = Column(Integer, primary_key=True)
company_id = Column(Integer, ForeignKey('company.id'), nullable=False)
stakeholder_id = Column(Integer, ForeignKey('company.id'), nullable=False)
company = relationship("Company", foreign_keys='company_id')
stakeholder = relationship("Company", foreign_keys='stakeholder_id')
I have seen similar questions here but some of the answers recommend one uses a primaryjoin yet in the documentation it states that you don't need the primaryjoin in this situation.
Tried removing quotes from the foreign_keys and making them a list. From official documentation on Relationship Configuration: Handling Multiple Join Paths
Changed in version 0.8: relationship() can resolve ambiguity between
foreign key targets on the basis of the foreign_keys argument alone;
the primaryjoin argument is no longer needed in this situation.
Self-contained code below works with sqlalchemy>=0.9:
from sqlalchemy import create_engine, Column, Integer, String, ForeignKey
from sqlalchemy.orm import relationship, scoped_session, sessionmaker
from sqlalchemy.ext.declarative import declarative_base
engine = create_engine(u'sqlite:///:memory:', echo=True)
session = scoped_session(sessionmaker(bind=engine))
Base = declarative_base()
#The business case here is that a company can be a stakeholder in another company.
class Company(Base):
__tablename__ = 'company'
id = Column(Integer, primary_key=True)
name = Column(String(50), nullable=False)
class Stakeholder(Base):
__tablename__ = 'stakeholder'
id = Column(Integer, primary_key=True)
company_id = Column(Integer, ForeignKey('company.id'), nullable=False)
stakeholder_id = Column(Integer, ForeignKey('company.id'), nullable=False)
company = relationship("Company", foreign_keys=[company_id])
stakeholder = relationship("Company", foreign_keys=[stakeholder_id])
Base.metadata.create_all(engine)
# simple query test
q1 = session.query(Company).all()
q2 = session.query(Stakeholder).all()
The latest documentation:
http://docs.sqlalchemy.org/en/latest/orm/join_conditions.html#handling-multiple-join-paths
The form of foreign_keys= in the documentation produces a NameError, not sure how it is expected to work when the class hasn't been created yet. With some hacking I was able to succeed with this:
company_id = Column(Integer, ForeignKey('company.id'), nullable=False)
company = relationship("Company", foreign_keys='Stakeholder.company_id')
stakeholder_id = Column(Integer, ForeignKey('company.id'), nullable=False)
stakeholder = relationship("Company",
foreign_keys='Stakeholder.stakeholder_id')
In other words:
… foreign_keys='CurrentClass.thing_id')