python - database save How to store a dictionary on a Django Model?

5 Answers

If you don't need to query by any of this extra data, then you can store it as a serialized dictionary. Use repr to turn the dictionary into a string, and eval to turn the string back into a dictionary. Take care with eval that there's no user data in the dictionary, or use a safe_eval implementation.

django models dictionary

I need to store some data in a Django model. These data are not equal to all instances of the model.

At first I thought about subclassing the model, but I’m trying to keep the application flexible. If I use subclasses, I’ll need to create a whole class each time I need a new kind of object, and that’s no good. I’ll also end up with a lot of subclasses only to store a pair of extra fields.

I really feel that a dictionary would be the best approach, but there’s nothing in the Django documentation about storing a dictionary in a Django model (or I can’t find it).

Any clues?

Another clean and fast solution can be found here:

For convenience I copied the simple instructions.


pip install jsonfield


from django.db import models
from jsonfield import JSONField

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

I'm not sure exactly sure of the nature of the problem you're trying to solve, but it sounds curiously similar to Google App Engine's BigTable Expando.

Expandos allow you to specify and store additional fields on an database-backed object instance at runtime. To quote from the docs:

import datetime
from google.appengine.ext import db

class Song(db.Expando):
  title = db.StringProperty()

crazy = Song(title='Crazy like a diamond',
             author='Lucy Sky',

crazy.last_minute_note=db.Text('Get a train to the station.')

Google App Engine currently supports both Python and the Django framework. Might be worth looking into if this is the best way to express your models.

Traditional relational database models don't have this kind of column-addition flexibility. If your datatypes are simple enough you could break from traditional RDBMS philosophy and hack values into a single column via serialization as @Ned Batchelder proposes; however, if you have to use an RDBMS, Django model inheritance is probably the way to go. Notably, it will create a one-to-one foreign key relation for each level of derivation.

Being "not equal to all instances of the model" sounds to me like a good match for a "Schema-free database". CouchDB is the poster child for that approach and you might consider that.

In a project I moved several tables which never played very nice with the Django ORM over to CouchDB and I'm quite happy with that. I use couchdb-python without any of the Django-specific CouchDB modules. A description of the data model can be found here. The movement from five "models" in Django to 3 "models" in Django and one CouchDB "database" actually slightly reduced the total lines of code in my application.