How to store a dictionary in a Django database model's field

Question

I need to save a dictionary in a model's field. How do I do that?

For example I have this code:

def create_random_bill(self):
    name_chars = re.compile("[a-zA-Z0-9 -_]")
    bill_name = "".join(random.choice(name_chars for x in range(10)))
    rand_products = random.randint(1,100)
    for x in rand_products:
        bill_products = 
    new_bill = Bill.new(name=bill_name, date=datetime.date, products=bill_products)
    new_bill.save()

What do I write for "bill_products=" so it saves some random products, from my Product model to this bill?

This is the bill's model description:

class Bill(models.Model):
    name = models.CharField(max_length=255)
    date = models.DateTimeField(auto_now_add=True)
    products = models.ManyToManyField(Product, related_name="bills")

And also the product's model description:

class Product(models.Model):
    name = models.CharField(max_length=255)
    price = models.IntegerField()

If there's anything else i should add just leave a comment. Thanks!

What the heck is a "python database model"? Are you using a particular ORM or framework? this looks a bit 'django-ish' — SingleNegationElimination, Mar 13 '12 at 14:58
Yes it is Django. I'm just starting to work with it so I might mix up "Python" with "Django". I'll change the title. — Radu Gheorghiu, Mar 13 '12 at 15:02

score 21 · Answer 1 · answered May 08 '13 at 09:54

21

I just discovered the django-jsonfield package, which

is a reusable Django field that allows you to store validated JSON in your model.

Looks like a viable option to achieve what you want.

answered May 08 '13 at 09:54

ramiro

878
9
20

2

Thanks for the update. It's good to know, even if this issue has been fixed a long time ago. – Radu Gheorghiu May 08 '13 at 12:09
1

This repo is not maintained since 2017. https://github.com/adamchainz/django-jsonfield is a fork of 2nd solution. Also, PostGres supports JSON fields natively: https://docs.djangoproject.com/en/dev/ref/contrib/postgres/fields/#jsonfield – Kangur Jul 24 '19 at 09:49

score 10 · Answer 2 · answered Jan 24 '17 at 21:07

Using a custom field type is my preferred solution - I'd rather have a few lines of custom code than support an entire 3rd party library for a single field type. Tony Abou-Assaleh has a great solution, but won't work for newer versions of Django.

This is verified to work with Django 1.10.4

import json

from django.db import models
from django.core.serializers.json import DjangoJSONEncoder


class JSONField(models.TextField):
    """
    JSONField is a generic textfield that neatly serializes/unserializes
    JSON objects seamlessly.
    Django snippet #1478

    example:
        class Page(models.Model):
            data = JSONField(blank=True, null=True)


        page = Page.objects.get(pk=5)
        page.data = {'title': 'test', 'type': 3}
        page.save()
    """

    def to_python(self, value):
        if value == "":
            return None

        try:
            if isinstance(value, str):
                return json.loads(value)
        except ValueError:
            pass
        return value

    def from_db_value(self, value, *args):
        return self.to_python(value)

    def get_db_prep_save(self, value, *args, **kwargs):
        if value == "":
            return None
        if isinstance(value, dict):
            value = json.dumps(value, cls=DjangoJSONEncoder)
        return value

This does not handle serialization (`manage.py dumpdata`), does not handle lookups in db, does not handle editing in Django adming (because representation shown is that of serialized dict), etc. Maybe in the end using a 3rd party lib saves you from running into errors. — Kangur, Nov 14 '19 at 16:23

score 10 · Answer 3 · answered Mar 13 '12 at 20:40

One convenient way to store a JSON representation in a model is to use a custom field type:

class JSONField(models.TextField):
    """
    JSONField is a generic textfield that neatly serializes/unserializes
    JSON objects seamlessly.
    Django snippet #1478

    example:
        class Page(models.Model):
            data = JSONField(blank=True, null=True)


        page = Page.objects.get(pk=5)
        page.data = {'title': 'test', 'type': 3}
        page.save()
    """

    __metaclass__ = models.SubfieldBase

    def to_python(self, value):
        if value == "":
            return None

        try:
            if isinstance(value, basestring):
                return json.loads(value)
        except ValueError:
            pass
        return value

    def get_db_prep_save(self, value, *args, **kwargs):
        if value == "":
            return None
        if isinstance(value, dict):
            value = json.dumps(value, cls=DjangoJSONEncoder)
        return super(JSONField, self).get_db_prep_save(value, *args, **kwargs)

I saved this utils/fields.py and in my model from utils.fields import JSONField. There are many more goodies in the django-annoying app, which is where this snippet came from.

score 6 · Accepted Answer · answered Mar 13 '12 at 15:12

6

Probably the cleanest thing to do would be to create another "Products" table and have a many-to-many relationship. (See here: https://docs.djangoproject.com/en/dev/topics/db/models/#many-to-many-relationships . In the docs they use the example of a pizza having many toppings.)

The other option would be to serialize your bill_products. In that case, you'd do something like:

bill_products = json.dumps([rand_products])

This would be outside of the for loop (although, in your example above, rand_products is only a single value, so you'll need to fix that).

answered Mar 13 '12 at 15:12

gdw2

7,558
4
46
49

If you could provide an example it would be very nice. I'm afraid I didn't understand how exactly to translate a many-to-many relationship to my code. Am I on the right idea path with saving an array with product names in that field? Or could I do it in a simpler approach? – Radu Gheorghiu Mar 14 '12 at 17:23
@reos You are on on the right track. Here is a full example: https://www.djangoproject.com/documentation/0_91/models/many_to_many/ – gdw2 Mar 14 '12 at 18:26

score 4 · Answer 5 · answered Jan 01 '15 at 14:25

4

If postgres is your backend, consider the hstore field which has native support from django

answered Jan 01 '15 at 14:25

wjin

954
1
8
13

Thank you for your answer, but this has long been answered and resolved. And the backend for this was SQLLite – Radu Gheorghiu Jan 01 '15 at 14:27

score 3 · Answer 6 · answered Dec 23 '13 at 05:41

3

You can use serialization/deserialization from pickle module:

http://docs.python.org/library/pickle.html

answered Dec 23 '13 at 05:41

Alejandro Veintimilla

10,743
23
91
180

score 2 · Answer 7 · answered Mar 13 '12 at 15:10

2

I think that I would create the field as models.CharField() and then encode the dictionary as a JSON string and save that string into the database. Then you can decode the JSON string back into a dictionary when you read it out.

answered Mar 13 '12 at 15:10

brian buck

3,284
3
23
24

I added the Bill's model description. – Radu Gheorghiu Mar 13 '12 at 15:14

score 1 · Answer 8 · answered Nov 14 '19 at 16:25

1

If using PostGres you can store it in natively supported JSON field: https://docs.djangoproject.com/en/dev/ref/contrib/postgres/fields/#jsonfield

Otherwise I'd recommend @ramiro answer with 3rd party lib https://stackoverflow.com/a/16437627/803174

answered Nov 14 '19 at 16:25

Kangur

7,823
3
30
32

score 0 · Answer 9 · answered Jun 29 '20 at 08:28

according to Django doc you can use :

from django.contrib.postgres.fields import JSONField
from django.db import models

class Dog(models.Model):
    name = models.CharField(max_length=200)
    data = JSONField()

    def __str__(self):
        return self.name

and create with this :

Dog.objects.create(name='Rufus', data={
     'breed': 'labrador',
     'owner': {
         'name': 'Bob',
         'other_pets': [{
             'name': 'Fishy',
         }],
     },
})

I hope this could help you.

How to store a dictionary in a Django database model's field

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