How to autogenerate UUID for Postgres in Python?

Question

I'm trying to create objects in Postgres db.

I'm using this approach https://websauna.org/docs/narrative/modelling/models.html#uuid-primary-keys

class Role(Base):
    __tablename__ = 'role'

    # Pass `binary=False` to fallback to CHAR instead of BINARY
    id = sa.Column(UUIDType(binary=False), primary_key=True)

But when I create object

user_role = Role(name='User')
db.session.add(user_role)
db.session.commit()

I have the following error:

sqlalchemy.exc.IntegrityError: (psycopg2.IntegrityError) null value in column "id" violates not-null constraint

Looks like I didn't provide any ID. So, how I can make the database auto-generate it or generate on my own?

Possible duplicate of Flask-Sqlalchemy, Primary key for secondary table in many-to-many relationship — mimic
– mimic, Commented Aug 28, 2018 at 23:13

Schwern · Accepted Answer · 2018-08-29 06:56:59Z

33

You appear to be using this code. It's missing a default for the column. You're emulating this SQL:

id UUID PRIMARY KEY DEFAULT uuid_generate_v4()

But you've already linked to the correct code.

id = Column(UUID(as_uuid=True),
    primary_key=True,
    server_default=sqlalchemy.text("uuid_generate_v4()"),)

Alternatively if you don't want to load a Postgres UUID extension, you can create the UUIDs in Python.

from uuid import uuid4

id = Column(UUID(as_uuid=True),
    primary_key=True,
    default=uuid4,)

answered Aug 29, 2018 at 6:56

Schwern

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1 Comment

xlm Over a year ago

For Postgres 13+ we no longer need to use the uuid-ossp extension and can instead use the built-in gen_random_uuid()

benvc · Accepted Answer · 2018-08-29 15:05:06Z

5

You could use the uuid module and just set a column default. For example:

from uuid import uuid4
from sqlalchemy import Column, String

class Role(Base):
    __tablename__ = 'role'

    id = Column(String, primary_key=True, default=uuid4)

edited Aug 29, 2018 at 15:05

answered Aug 28, 2018 at 23:20

benvc

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1 Comment

benvc Over a year ago

If you are looking to use the PostgreSQL UUID type (which I believe has some performance advantages), see the answer from @Schwern.

mimic · Accepted Answer · 2020-01-23 17:17:44Z

2

What I actually came to is:

import uuid

class SomeClass(db.Model):
    __tablename__ = 'someclass'

    id = db.Column(UUID(as_uuid=True),
        primary_key=True, default=lambda: uuid.uuid4().hex)

answered Jan 23, 2020 at 17:17

mimic

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1 Comment

bfontaine Over a year ago

Downvoted because this is confusing: you explicitely say you want a UUID type with as_uuid=True, but you provide a default function that returns an str.

Jérémy DEMANGE · Accepted Answer · 2021-02-17 17:59:30Z

-4

import uuid

myId = uuid.uuid4()
print(myId)

answered Feb 17, 2021 at 17:59

Jérémy DEMANGE

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