Hi all,
when I try to create a django app in a specified folder with the same name
as the app I get the following error:
*CommandError: 'my_app' conflicts with the name of an existing Python
module and cannot be used as an app name. Please try another name.*
These are the commands I'm using:
$
increased it to 50 and I'm getting 50 hits.
On Thu, May 25, 2017 at 12:06 AM, Nick Gilmour <nickefo...@gmail.com> wrote:
> Thanks for the hint!
> It seems to be related with Django's pagination:
> with:
> {{ page.paginator.count }}
> and
> {{ page.paginator.num_pages }}
&g
works...
On Wed, May 24, 2017 at 4:55 PM, Avraham Serour <tovm...@gmail.com> wrote:
> it sounds like elastic is paginating, did you check that?
>
> On Tue, May 23, 2017 at 12:41 AM, Nick Gilmour <nickefo...@gmail.com>
> wrote:
>
>> Hi all,
>>
>>
&
Hi all,
I'm following an example to setup Django with Haystack and ES from here:
https://krzysztofzuraw.com/blog/2016/haystack-elasticsearch-part-one.html
Everything seems to be working OK except from the number of the results –
they are always maximal 20. But actually a single query should
>
> Neither one of these is what was being asked for.
I didn't say I have found what I was looking for.
All answers are sufficient to me for now. Thanks.
On Sun, May 14, 2017 at 10:51 PM, James Bennett <ubernost...@gmail.com>
wrote:
> On Sun, May 14, 2017 at 1:36 PM, Nick G
Definition of url:
*def url(regex, view, kwargs=None, name=None):*
I also found this:
*def view(request, *args, **kwargs):*
here:
*.../django/views/generic/base.py*
On Sat, May 13, 2017 at 11:15 PM, Daniel Roseman
wrote:
> On Friday, 12 May 2017 15:11:55 UTC+1,
Thanks for the link. I understand these examples but still I would like to
see how a view function is defined.
I also would like to see it in PyCharm. When I click on a view function in
PyCharm I only see:
def myview_function(request)
Inferred type: (request: Any) -> HttpResponse
I would expect
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