AttributeError: 'InstrumentedList' object has no attribute

Question

I have these tables tables:

class Thing(Base):
    __tablename__ = 'thing'
    id = Column(Integer, primary_key=True)

class User(Base):
    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)

class Voteinfo(Base):
    __tablename__ = 'voteinfo'
    thing_id = Column(Integer, ForeignKey('thing.id'), primary_key=True)
    thing = relationship('Thing', backref='voteinfo')
    upvotes = Column(Integer)
    downvotes = Column(Integer)

    def __init__(self, thing)
        self.thing = thing

class VoteThing(Base):
    __tablename__ = 'votething'
    id = Column(Integer, primary_key=True)
    voter_id = Column(Integer, ForeignKey('voter.id'))
    voter = relationship('Voter', backref='votescast')
    thing_id = Column(Integer, ForeignKey('thing.id'))
    thing = relationship('Thing', backref='votesreceived')
    value = Column(Boolean)

    def __init__(self, voter, thing, value):
        if value is True:
            thing.voteinfo.upvotes += 1
        else:
            thing.voteinfo.downvotes += 1

When I try to run this, I get this error code in the "if value is True" clause:

AttributeError: 'InstrumentedList' object has no attribute 'upvotes'

I've tried giving Voteinfo its own unique ID and adding uselist=False to the relationship. I've tried replacing the relationship to thing from VoteThing to Voteinfo, but that didn't help either. I don't know what an InstrumentedList is. What is going on?

since `thing` is a parameter to `__init__`, presumably you're passing it when you instantiate the VoteThing. So what are you passing? — Daniel Roseman, Oct 06 '11 at 08:59
Yes, I am passing a thing: thing1 = Thing(), user1 = User(), voteinfo1 = Voteinfo(thing1), votething1 = VoteThing(user1, thing1, True) — Jonathan Ong, Oct 06 '11 at 09:06

score 32 · Accepted Answer · edited Jun 17 '18 at 05:27

32

As explained in the documentation, here : https://docs.sqlalchemy.org/en/latest/orm/basic_relationships.html#one-to-one, you have to add uselist=False not to the relationship, but to the backref.

thing = relationship('Thing', backref=backref('voteinfo', uselist=False))

edited Jun 17 '18 at 05:27

dangel

7,238
7
48
74

answered Oct 06 '11 at 09:13

madjar

12,691
2
44
52

2

For those of you reading this helpful answer from The Future, backref() is now a legacy feature: https://docs.sqlalchemy.org/en/20/orm/backref.html. Looks like uselist goes on the "many" side of the relationship marker: https://docs.sqlalchemy.org/en/20/orm/basic_relationships.html#setting-uselist-false-for-non-annotated-configurations – Kirk Kittell Jan 06 '23 at 14:43
Careful here, using `uselist=False` when a collection is expected will most likely result in unexpected behavior. I.e. it may seem to work ok but really it provides a single item rather than the whole collection. – EliadL Jul 31 '23 at 13:28

score 0 · Answer 2 · answered Feb 08 '22 at 10:14

It may been useful to add lazy='dynamic' argument to request that the query is not automatically executed and of course refresh (thing = relationship('Thing', backref='voteinfo', lazy='dynamic')), otherwise expression is issued internally call all() to return the list of 'thing' hence AttributeError: 'InstrumentedList' object has no attribute error!!

score 0 · Answer 3 · answered Feb 17 '23 at 20:12

crazy how things have changed but this helped a ton!, an error raised when i used lazy = 'dynamic' but this is what my code looks like

tool = db.relationship("Tool", backref="activities", uselist=False)

it gives me db.model type and it's what i was looking for. Instead of a instrumented class list type.

AttributeError: 'InstrumentedList' object has no attribute

3 Answers3

Linked