I think Django documentation is well written
On Wed, Sep 14, 2016 at 11:52 PM, Timothy Steele <
steeletimothy2...@gmail.com> wrote:
> Thanks Tim it has work this time but can any one help me with a beginner
> book to learn Django please
>
>
>
> On Wednesday, September 14, 2016 at 4:44:14 PM UTC+1
Have you looked at the official Django documentation?
http://masteringdjango.com/django-book/ is another nice resource.
On Wednesday, September 14, 2016 at 2:22:18 PM UTC-4, Timothy Steele wrote:
>
> Thanks Tim it has work this time but can any one help me with a beginner
> book to learn Django
Thanks Tim it has work this time but can any one help me with a beginner
book to learn Django please
On Wednesday, September 14, 2016 at 4:44:14 PM UTC+1, Timothy Steele wrote:
>
> Instead of providing the view itself, i want to pass its module path as a
> string in the urls.py file as the c
Tim Graham, I did it but i get back this error *login() missing 1 required
positional argument: 'user' *
On Wednesday, September 14, 2016 at 4:44:14 PM UTC+1, Timothy Steele wrote:
>
> Instead of providing the view itself, i want to pass its module path as a
> string in the urls.py file as the
On 14 September 2016 at 17:44, Timothy Steele
wrote:
> Instead of providing the view itself, i want to pass its module path as a
> string in the urls.py file as the code below
>
> urlpatterns = ['',
> url(r'^$', main_page),
> url(r'^user/(\w+)/$', user_page),
> url(r'^login/$', 'd
Import the view: from django.contrib.auth.views import login
and replace 'django.contrib.auth.views.login' with a reference to the login
function.
from django.contrib.auth.views import login
urlpatterns = [ <-- remove the empty string on this line
...
url(r'^login/$', login),
]
On Wed
Instead of providing the view itself, i want to pass its module path as a
string in the urls.py file as the code below
urlpatterns = ['',
url(r'^$', main_page),
url(r'^user/(\w+)/$', user_page),
url(r'^login/$', 'django.contrib.auth.views.login'),
]
but it return this informati
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