Error when creating a django app in a specified folder with the same name as the app

[Nick Gilmour]

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:
$ mkdir backend/apps/my_app
$ python manage startapp my_app ./backend/apps/my_app

Is this normal? Any ideas why is this happening?

Regards,

Nick

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Re: Django+Haystack+Elastic issue

[Nick Gilmour]

It is related with Django's pagination...

The results per page value can be set with:
HAYSTACK_SEARCH_RESULTS_PER_PAGE = 20
in the settings.py.

I had found and defined this setting before but it wasn't working because I
had this in the HAYSTACK_CONNECTIONS dict (don't ask me why...).

I have 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 }}
> I can see that I have 58 hits and 3 pages, which is correct.
> I was getting maximal 2 pages with 20 pages each.
> Now I have to check how pagination in Django 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,
>>>
>>>
>>>
>>> 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 give more
>>> than 50 results.
>>>
>>> So, where does this limitation come from? How can I change it?
>>>
>>> I cannot find a setting neither for haystack nor for django which
>>> prevents showing more than 20 results.
>>>
>>>
>>>
>>> Thanks,
>>>
>>> Nick
>>>
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Re: Django+Haystack+Elastic issue

[Nick Gilmour]

Thanks for the hint!
It seems to be related with Django's pagination:
with:
{{ page.paginator.count }}
and
{{ page.paginator.num_pages }}
I can see that I have 58 hits and 3 pages, which is correct.
I was getting maximal 2 pages with 20 pages each.
Now I have to check how pagination in Django 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,
>>
>>
>>
>> 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 give more
>> than 50 results.
>>
>> So, where does this limitation come from? How can I change it?
>>
>> I cannot find a setting neither for haystack nor for django which
>> prevents showing more than 20 results.
>>
>>
>>
>> Thanks,
>>
>> Nick
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Django users" group.
>> To unsubscribe from this group and stop receiving emails from it, send an
>> email to django-users+unsubscr...@googlegroups.com.
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>> Visit this group at https://groups.google.com/group/django-users.
>> To view this discussion on the web visit https://groups.google.com/d/ms
>> gid/django-users/CAH-drozT6m%2Be_KX4Kni_U5nJXQp8NF8EeXb9cJo
>> 5OOhtUt1LoQ%40mail.gmail.com
>> <https://groups.google.com/d/msgid/django-users/CAH-drozT6m%2Be_KX4Kni_U5nJXQp8NF8EeXb9cJo5OOhtUt1LoQ%40mail.gmail.com?utm_medium=email_source=footer>
>> .
>> For more options, visit https://groups.google.com/d/optout.
>>
>
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>

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Django+Haystack+Elastic issue

[Nick Gilmour]

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 give more
than 50 results.

So, where does this limitation come from? How can I change it?

I cannot find a setting neither for haystack nor for django which prevents
showing more than 20 results.



Thanks,

Nick

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Re: View Function Signature?

[Nick Gilmour]

>
> 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 Gilmour <nickefo...@gmail.com>
> wrote:
>
>> Definition of url:
>>
>> *def url(regex, view, kwargs=None, name=None):*
>>
>> I also found this:
>> *def view(request, *args, **kwargs):*
>>
>>
> Neither one of these is what was being asked for.
>
> And like several people have said, what's being asked for doesn't exist: a
> Django view is defined as a callable Python object which takes as its first
> position argument an HttpRequest object, may or may not accept additional
> positional and/or keyword arguments, and either returns an HttpResponse
> object or raises an exception. That definition does not allow for a single
> universal argument signature that will work on all views.
>
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> .
>
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>

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Re: View Function Signature?

[Nick Gilmour]

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, nickeforos wrote:
>>
>> 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 to see something like this:
>> view_function(request, *args, **kwargs)
>>
>> I've also looked in django's source but I cannot locate it.
>>
>
>
> But this isn't how Python works. You don't need a defined signature or
> prototype for a class of functions. A view is simply called with the
> request and any parameters passed from the URL, that's all there is to it.
> --
> DR.
>
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Re: View Function Signature?

[Nick Gilmour]

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 to see something like this:
view_function(request, *args, **kwargs)

I've also looked in django's source but I cannot locate it.

On Fri, May 12, 2017 at 3:05 PM, nickeforos  wrote:

> I'm following the django tutorials and documentaion as well as the django
> book.
> In
> https://docs.djangoproject.com/en/1.11/topics/http/views/
> there is this view function:
> def current_datetime(request):
> now = datetime.datetime.now()
> html = "It is now %s." % now
> return HttpResponse(html)
>
> and in:
> http://djangobook.com/django-views-dynamic-content/http://
> djangobook.com/django-views-dynamic-content/
> there is this view function:
> def hours_ahead(request, offset):
> try:
> offset = int(offset)
> except ValueError:
> raise Http404()
> dt = datetime.datetime.now() + datetime.timedelta(hours=offset)
> html = "In %s hour(s), it will be  %s." % (offset, dt)
> return HttpResponse(html)
>
> So in the view function I can pass *request* as well as an *argument*
> from the url but how does the signature of a view function look like?
> Where can I find it in the documentation?
> Is it possible to see such signatures in PyCharm?
>
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