'django update_or_create gets "duplicate key value violates unique constraint "

Maybe I misunderstand the purpose of Django's update_or_create Model method.

Here is my Model:

from django.db import models
import datetime
from vc.models import Cluster

class Vmt(models.Model):
    added = models.DateField(default=datetime.date.today, blank=True, null=True)
    creation_time = models.TextField(blank=True, null=True)
    current_pm_active = models.TextField(blank=True, null=True)     
    current_pm_total = models.TextField(blank=True, null=True)
    ... more simple fields ...
    cluster = models.ForeignKey(Cluster, null=True)


    class Meta:
        unique_together = (("cluster", "added"),)

Here is my test:

from django.test import TestCase
from .models import *
from vc.models import Cluster
from django.db import transaction


# Create your tests here.
class VmtModelTests(TestCase):
    def test_insert_into_VmtModel(self):
        count = Vmt.objects.count()
        self.assertEqual(count, 0)

        # create a Cluster
        c = Cluster.objects.create(name='test-cluster')
        Vmt.objects.create(
            cluster=c,
            creation_time='test creaetion time',
            current_pm_active=5,
            current_pm_total=5,
            ... more simple fields ...
        )
        count = Vmt.objects.count()
        self.assertEqual(count, 1)
        self.assertEqual('5', c.vmt_set.all()[0].current_pm_active)

        # let's test that we cannot add that same record again
        try:
            with transaction.atomic():

                Vmt.objects.create(
                    cluster=c,
                    creation_time='test creaetion time',
                    current_pm_active=5,
                    current_pm_total=5,
                    ... more simple fields ...
                )
                self.fail(msg="Should violated integrity constraint!")
        except Exception as ex:
            template = "An exception of type {0} occurred. Arguments:\n{1!r}"
            message = template.format(type(ex).__name__, ex.args)
            self.assertEqual("An exception of type IntegrityError occurred.", message[:45])

        Vmt.objects.update_or_create(
            cluster=c,
            creation_time='test creaetion time',
            # notice we are updating current_pm_active to 6
            current_pm_active=6,
            current_pm_total=5,
            ... more simple fields ...
        )
        count = Vmt.objects.count()
        self.assertEqual(count, 1)

On the last update_or_create call I get this error:

IntegrityError: duplicate key value violates unique constraint "vmt_vmt_cluster_id_added_c2052322_uniq"
DETAIL:  Key (cluster_id, added)=(1, 2018-06-18) already exists.

Why didn't wasn't the model updated? Why did Django try to create a new record that violated the unique constraint?



Solution 1:[1]

The update_or_create(defaults=None, **kwargs) has basically two parts:

  1. the **kwargs which specify the "filter" criteria to determine if such object is already present; and
  2. the defaults which is a dictionary that contains the fields mapped to values that should be used when we create a new row (in case the filtering fails to find a row), or which values should be updated (in case we find such row).

The problem here is that you make your filters too restrictive: you add several filters, and as a result the database does not find such row. So what happens? The database then aims to create the row with these filter values (and since defaults is missing, no extra values are added). But then it turns out that we create a row, and that the combination of the cluster and added already exists. Hence the database refuses to add this row.

So this line:

Model.objects.update_or_create(field1=val1,
                               field2=val2,
                               defaults={
                                   'field3': val3,
                                   'field4': val4
                               })

Is to semantically approximately equal to:

try:
    item = Model.objects.get(field1=val1, field2=val2)
except Model.DoesNotExist:
    Model.objects.create(field1=val1, field2=val2, field3=val3, field4=val4)
else:
    item = Model.objects.filter(
        field1=val1,
        field2=val2,
    ).update(
        field3 = val3
        field4 = val4
    )

(but the original call is typically done in a single query).

You probably thus should write:

Vmt.objects.update_or_create(
    cluster=c,
    creation_time='test creaetion time',
    defaults = {        
        'current_pm_active': 6,
        'current_pm_total': 5,
    }
)

(or something similar)

Solution 2:[2]

You should separate your field:

  1. Fields that should be searched for
  2. Fields that should be updated

for example: If I have the model:

class User(models.Model):
    username = models.CharField(max_length=200)
    nickname = models.CharField(max_length=200)

And I want to search for username = 'Nikolas' and update this instance nickname to 'Nik'(if no User with username 'Nikolas' I need to create it) I should write this code:

User.objects.update_or_create(
    username='Nikolas', 
    defaults={'nickname': 'Nik'},
)

see in https://docs.djangoproject.com/en/3.1/ref/models/querysets/

Solution 3:[3]

This is already answered well in the above.

To be more clear the update_or_create() method should have **kwargs as those parameters on which you want to check if that data already exists in DB by filtering.

select some_column from table_name where column1='' and column2='';

Filtering by **kwargs will give you objects. Now if you wish to update any data/column of those filtered objects, you should pass them in defaults param in update_or_create() method.

so lets say you found an object based on a filter now the default param values are expected to be picked and updated.

and if there's no matching object found based on the filter then it goes ahead and creates an entry with filters and the default param passed.

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Willem Van Onsem
Solution 2 Simone Pozzoli
Solution 3 Rohit sai