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Django JSONField

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Elias Imokhai answered on September 3, 2020 Popularity 10/10 Helpfulness 2/10

answer Django JSONField

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Django JSONField

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from django.db import models

class Patient(models.Model):
    name = models.CharField(max_length=256)
    data = models.JSONField()


# Create a patient instance
Patient(name='John Doe', data={
    'address': '123 Some House Number', 
    'city': 'Some City',
    'state': 'Utah',
})

# Filter and delete the patient
Patient.objects.filter(patient='John Doe', data__state='Utah')

Popularity 10/10 Helpfulness 2/10 Language css

Source: oak-tree.tech

Tags: css django-jsonfield

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Contributed on Apr 11 2022

Elias Imokhai

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django json field

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# New version
class MyModel(models.Model):
  attribute = models.JSONField(...)

  
  
# Old version (<3.1, Postgres only)
from django.contrib.postgres.fields import JSONField

class MyModel(models.Model):
  attribute = JSONField(...)

Popularity 10/10 Helpfulness 4/10 Language python

Source: docs.djangoproject.com

Tags: django dj

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Contributed on Jul 18 2021

Thankful Teira

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jsonfield django

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>>> Dog.objects.create(name='Rufus', data={
...     'breed': 'labrador',
...     'owner': {
...         'name': 'Bob',
...         'other_pets': [{
...             'name': 'Fishy',
...         }],
...     },
... })
>>> Dog.objects.create(name='Meg', data={'breed': 'collie', 'owner': None})

>>> Dog.objects.filter(data__breed='collie')
<QuerySet [<Dog: Meg>]>

Popularity 10/10 Helpfulness 2/10 Language python

Source: docs.djangoproject.com

Tags: django dj

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Contributed on Sep 03 2020

Dead Donkey

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Django JSONField

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