Process Django JSONField as YAML

Question

In Django currently Postgres JSONField model field is shown as plain JSON text in CharField form field.

I'd like to present data in CharField as YAML text (while keeping it in JSON format internally) and while saving convert it back to JSON like:

yaml.dump(json.loads(value))

Current:

Necessary:

How to accomplish that?

Thanks.

Answer 1

Solution is to subclass JSONField (both model and form) and just replace json with yaml. Made as separate package: https://github.com/mike-tk/django-yamlfield

from django.contrib.postgres import fields, forms
import yaml
from django.utils.translation import gettext_lazy as _

class InvalidYAMLInput(str):
    pass


class YAMLString(str):
    pass


class YAMLFormField(forms.JSONField):
    default_error_messages = {
        'invalid': _("'%(value)s' value must be valid YAML."),
    }

    def to_python(self, value):
        if self.disabled:
            return value
        if value in self.empty_values:
            return None
        elif isinstance(value, (list, dict, int, float, YAMLString)):
            return value
        try:
            converted = yaml.load(value)
        except yaml.YAMLError:
            raise forms.ValidationError(
                self.error_messages['invalid'],
                code='invalid',
                params={'value': value},
            )
        if isinstance(converted, str):
            return YAMLString(converted)
        else:
            return converted

    def bound_data(self, data, initial):
        if self.disabled:
            return initial
        try:
            return yaml.load(data)
        except yaml.YAMLError:
            return InvalidYAMLInput(data)

    def prepare_value(self, value):
        if isinstance(value, InvalidYAMLInput):
            return value
        return yaml.dump(value, default_flow_style=False)


class YAMLField(fields.JSONField):
    def formfield(self, **kwargs):
        defaults = {'form_class': YAMLFormField}
        defaults.update(kwargs)
        return super().formfield(**defaults)