How to support JSONField

[lucas-bremond]

I've tried the following:

from django.contrib.postgres.fields import JSONField
from fernet_fields import EncryptedField

class EncryptedJSONField (EncryptedField, JSONField):
    ...

private_column = EncryptedJSONField(default = dict)

but got the following error:

django.db.utils.ProgrammingError: column "private_column" is of type jsonb but expression is of type bytea

Any suggestion?

[marcus-campos]

@lucas-bremond I managed to solve this way, I don't know if it's the best way, but it worked for me

from fernet_fields import EncryptedField

from django.contrib.postgres.fields import JSONField


class CustomJSONField(JSONField):
    def db_type(self, connection):
        return 'text'


class JSONEncryptedField(EncryptedField, CustomJSONField):
    pass