Re: TypeError: view must be a callable or a list/tuple in the case of include(). in urls.py

[James Schneider]

On Sep 5, 2016 6:06 AM, "arun kumar"  wrote:
>
> Hi,
>
>I'm working in Django 1.10 and I got the type error  in the below file
at this line " url(r'^$', views.dashboard, name = 'dashboard'),"
>
> from django.conf.urls import include, url
> from . import views
>
> urlpatterns = [
> #preview login view
> #url(r'^login/$',views.user_login, name='login')
>
> # logged - in user dashboard
> url(r'^$', views.dashboard, name = 'dashboard'),
>
> #login / logout urls
> url(r'^login/$', 'django.contrib.auth.views.login', name='login'),
>

Here is where your problem starts. Django 1.10 removed the ability to use
string arguments to refer to views. See the deprecation notice in 1.8:

https://docs.djangoproject.com/en/1.8/ref/urls/#django.conf.urls.url

Import your views and pass them along directly.

-James

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Re: TypeError: view must be a callable or a list/tuple in the case of include(). in urls.py

['Abraham Varricatt' via Django users]

Not sure what the issue is without more info about the error, but is the 
regular expression for that view correct? Assuming that you want to divert 
folks to the root of your site can you try the following?

url(r'^/$', views.dashboard, name = 'dashboard')


NOTE: if the error still persists, it will help if you include the full 
traceback in your email.

Yours,
Abraham V.



On Monday, September 5, 2016 at 6:36:37 PM UTC+5:30, arun kumar wrote:
>
> Hi,
>
>I'm working in Django 1.10 and I got the type error  in the below file 
> at this line "* url(r'^$', views.dashboard, name = 'dashboard'),*"
>
> *from django.conf.urls import include, url*
> *from . import views*
>
> *urlpatterns = [*
> *#preview login view*
> *#url(r'^login/$',views.user_login, name='login')*
>
> *# logged - in user dashboard*
> *url(r'^$', views.dashboard, name = 'dashboard'),*
>
> *#login / logout urls*
> *url(r'^login/$', 'django.contrib.auth.views.login', name='login'),*
>
> *url(r'^logout/$', 'django.contrib.auth.views.logout', name='logout'),*
>
> *
> url(r'^logout-then-login/$','django.contrib.auth.views.logout_then_login', 
> name='logout_then_login')*
>
> *]*
>
> I'm new to Django and stuck because of this 
>

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TypeError: view must be a callable or a list/tuple in the case of include(). in urls.py

[arun kumar]

Hi,

   I'm working in Django 1.10 and I got the type error  in the below file 
at this line "* url(r'^$', views.dashboard, name = 'dashboard'),*"

*from django.conf.urls import include, url*
*from . import views*

*urlpatterns = [*
*#preview login view*
*#url(r'^login/$',views.user_login, name='login')*

*# logged - in user dashboard*
*url(r'^$', views.dashboard, name = 'dashboard'),*

*#login / logout urls*
*url(r'^login/$', 'django.contrib.auth.views.login', name='login'),*

*url(r'^logout/$', 'django.contrib.auth.views.logout', name='logout'),*

*
url(r'^logout-then-login/$','django.contrib.auth.views.logout_then_login', 
name='logout_then_login')*

*]*

I'm new to Django and stuck because of this 

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