django-form-admin 0.5.1

Option 1

In your form:

class FooForm(forms.Form, FormAdminDjango):
    pass

class FooModelForm(forms.ModelForm, FormAdminDjango):
    pass

In your template:

{{ form.as_django_admin }}

Option 2

But you don’t need inherit of FormAdminDjango, you may also do

In your form:

class FooForm(forms.Form):
    ...

    def as_django_admin(self):
        from formadmin.forms import as_django_admin
        return as_django_admin(self)

In your template:

{{ form.as_django_admin }}

Option 3

Or even without modify the form

Only in your template:

{% load formadmin_filters %}
{{ form|as_django_admin }}

0.5.1 (2018-03-05)

• Fix previous version (Django support > 1.8)

0.5.0 (2018-03-05)

• Django support > 1.8 (1.9 - 1.11)

• Improve readme file

• Thanks to:

  • bashu

  • predatell

0.4.2 (2013-10-28)

• Fix an error when to install this egg from pypi

0.4.1 (2013-10-04)

• Improvements in the example project

0.4.0 (2013-09-12)

• New way to render the form (filter)

• Update metainfo

• Example project

0.3.2 (2011-12-03)

• Subversion version