Django UpdateView sample

django_update_view.py

#views.py

from django.views.generic import UpdateView

class UpdateBook(UpdateView):

    model = Book
    form_class = BookForm
    template_name = 'create_form.html'
    success_url = '/books/'


#urls.py

from django.conf.urls import *
from library.views import UpdateBook

urlpatterns = patterns('',
    url('^update_book/(?P<pk>[\w-]+)$', UpdateBook.as_view(), name='update_book'),
)
# you MUST pass a pk or slug param to this view

thanks for your sharing, it's very helpful for me

Thank you flaviamissi, you saved me a lot of time.

If u wanna call this from some template, the template should have an url like this
{% for book in object_list %}
<a href="{% url "update_book" book.pk %}"> UPDATE {{ book.name }}

This will display a list of books ...and update options besides it

👍 Simple and to the point. Thanks

how to send context in success url?

Thank you! You are the best!

Thank..