Reputation: 27861
Here is a sample model:
class FooModel(models.Model):
foo = models.DecimalField(max_digits=6, decimal_places=3, null=True)
Serialize:
from django.core import serializers
obj = get_object_or_404(FooModel, pk=1)
data = serializers.serialize("json", [obj])
That will return something like:
[
{
"pk": 1,
"model": "app.foomodel",
"fields": {
"foo": "50"
}
}
]
Question
How can I make foo field to serialize as float, and not as string. I don't want to use float model type since float sometimes does not store decimals correctly.
Thank you in advance.
Upvotes: 2
Views: 4436
Reputation: 33716
If the value actually 50
vs. 50.0
? A Decimal
object maintains whatever was originally input, e.g. Decimal('50')
yields 50
.
>>> from decimal import Decimal
>>> d = Decimal('50')
>>> print d
50
Unfortunately the arguments provided to the DecimalField
constructor are only for restrictions on storing the values and not displaying them.
Now, because Django is just using the standard json/simplejson lib, you can specify a custom encoder when serializing, such as suggested in this question:
import decimal
import json
class DecimalEncoder(json.JSONEncoder):
def default(self, obj):
if isinstance(obj, decimal.Decimal):
return '%.2f' % obj # Display Decimal obj as float
return json.JSONEncoder.default(self, obj)
But it doesn't end there. As detailed in this blog post, Django is explicitly passing cls=DjangoJSONEncoder
to simplejson.dump(...)
, so we have to circumvent this by also creating a custom serializer object referencing the DecimalEncoder
we created:
from django.core.serializers.json import Serializer as JSONSerializer
class DecimalSerializer(JSONSerializer):
def end_serialization(self):
self.options.pop('stream', None)
self.options.pop('fields', None)
json.dump(self.objects, self.stream, cls=DecimalEncoder, **self.options)
Next, you instantiate DecimalSerializer
as your own serializer object, and magic happens:
my_serializer = DecimalSerializer()
print my_serializer.serialize([obj], indent=4)
Which yields:
[
{
"pk": 1,
"model": "app.foomodel",
"fields": {
"foo": "50.00"
}
}
]
This seems like a lot of work. It might just be easier to use Django's model validation to make sure that a FooModel.foo
field is always a floating point number, prior to saving. Here is brute force attempt at that:
from django.core.exceptions import ValidationError
class FooModel(models.Model):
foo = models.DecimalField(max_digits=6, decimal_places=3, null=True)
def clean(self):
if '.' not in str(self.foo):
raise ValidationError('Input must be float!')
def save(self, *args, **kwargs):
self.full_clean()
super(FooModel, self).save(*args, **kwargs)
And then:
>>> f = FooModel(foo='1')
>>> f.save()
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "/home/jathan/sandbox/foo/app/models.py", line 15, in save
self.full_clean()
File "/usr/local/lib/python2.6/dist-packages/django/db/models/base.py", line 828, in full_clean
raise ValidationError(errors)
ValidationError: {'__all__': [u'Input must be float!']}
Upvotes: 4