In the django docs, there's an example of using inlineformset_factory to edit already created objects
https://docs.djangoproject.com/en/dev/topics/forms/modelforms/#using-an-inline-formset-in-a-view
I changed the example to be this way:
def manage_books(request):
author = Author()
BookInlineFormSet = inlineformset_factory(Author, Book, fields=('title',))
if request.method == "POST":
formset = BookInlineFormSet(request.POST, request.FILES, instance=author)
if formset.is_valid():
formset.save()
return HttpResponseRedirect(author.get_absolute_url())
else:
formset = BookInlineFormSet(instance=author)
return render_to_response("manage_books.html", {
"formset": formset,
})
With the above, it renders only the inline model without the parent model.
To create a new object, say Author, with multiple Books associated to, using inlineformset_factory, what's the approach?
An example using the above Author Book model from django docs will be helpful. The django docs only provided example of how to edit already created object using inlineformset_factory but not to create new one
I am posting my final solutions, as per extensive assistant given by Onyeka.
Below I post the Add and Edit solutions of using inlineformset_factory of Django using the Author and Book example found in the Django Docs.
First, the Adding of Author object, with 3 extras of Book object to be appended.
Obviously, this goes into your views.py
This part goes into your urls.py, assuming views have been imported, and urlpatterns constructed already.
Now to the templates part. The edit Author object template (edit_author.html) looks like this (no styling applied)
To add a brand new Author object via template (add_author.html):
NOTE:
Using the action='.' might appear to be a cheap way of constructing the url, whereby the form posts the form data to the same page. With this example, using the action='.' for the edit_author.html template always got the form posted to /edit/ instead of /edit/1 or /edit/2
Constructing the url using the {% url 'edit_author' author_id %} ensures the form always posts to the right url. Failing to do use the {% url %} cost me lots of hours and trouble.
Big thanks to Onyeka.
I've done that using Django Class-Based Views.
Here's my approach:
models.py
forms.py
views.py
urls.py
manage_books.html
Notice:
inlineformset_factory
feature and Django generic Class-Based Viewsdjango-crispy-forms
is installed, and it's properly configured.I know it's more code that the showed solutions, but start to using Django Class-Based Views is great.
i did exactly what you are trying : https://github.com/yakoub/django_training/tree/master/article
you need to create a separate form using the prefix attribute . then when you save you need to iterate over all books and associate them with the author you just created .
I didn't read your question properly at first. You need to also render the the form for the parent model. I haven't tested this, I'm going off what I've done before and the previously linked answer, but it should work.
UPDATE
If you're using the view to both and edit, you should check for an Author ID first. If there's no ID, it'll render both forms as a new instance, whereas with an ID it'll, fill them with the existing data. Then you can check if there was a POST request.