Python 实用宝典

Question 1

It is unclear to me how it is best to access URL-parameters in class-based-views in Django 1.5.

Consider the following:

View:

from django.views.generic.base import TemplateView


class Yearly(TemplateView):
    template_name = "calendars/yearly.html"

    current_year = datetime.datetime.now().year
    current_month = datetime.datetime.now().month

    def get_context_data(self, **kwargs):
        context = super(Yearly, self).get_context_data(**kwargs)
        context['current_year'] = self.current_year
        context['current_month'] = self.current_month
        return context

URLCONF:

from .views import Yearly


urlpatterns = patterns('',
    url(
        regex=r'^(?P<year>\d+)/$',
        view=Yearly.as_view(),
        name='yearly-view'
    ),
)

I want to access the year parameter in my view, so I can do logic like:

month_names = [
    "January", "February", "March", "April", 
    "May", "June", "July", "August", 
    "September", "October", "November", "December"
]

for month, month_name in enumerate(month_names, start=1):
    is_current = False
    if year == current_year and month == current_month:
        is_current = True
        months.append({
            'month': month,
            'name': month_name,
            'is_current': is_current
        })

How would one best access the url parameter in CBVs like the above that is subclassed of TemplateView and where should one ideally place the logic like this, eg. in a method?

Question 2

To access the url parameters in class based views, use self.args or self.kwargs so you would access it by doing self.kwargs['year']

Question 3

In case you pass URL parameter like this:

http://<my_url>/?order_by=created

You can access it in class based view by using self.request.GET (its not presented in self.args nor in self.kwargs):

class MyClassBasedView(ObjectList):
    ...
    def get_queryset(self):
        order_by = self.request.GET.get('order_by') or '-created'
        qs = super(MyClassBasedView, self).get_queryset()
        return qs.order_by(order_by)

Question 4

I found this elegant solution, and for django 1.5 or higher, as pointed out here:

Django’s generic class based views now automatically include a view variable in the context. This variable points at your view object.

In your views.py:

from django.views.generic.base import TemplateView    

class Yearly(TemplateView):
    template_name = "calendars/yearly.html"
    # Not here 
    current_year = datetime.datetime.now().year
    current_month = datetime.datetime.now().month

    # dispatch is called when the class instance loads
    def dispatch(self, request, *args, **kwargs):
        self.year = kwargs.get('year', "any_default")

    # other code

    # needed to have an HttpResponse
    return super(Yearly, self).dispatch(request, *args, **kwargs)

The dispatch solution found in this question.
As the view is already passed within Template context, you don’t really need to worry about it. In your template file yearly.html, it is possible to access those view attributes simply by:

{{ view.year }}
{{ view.current_year }}
{{ view.current_month }}

You can keep your urlconf as it is.

It’s worth mentioning that getting information into your template’s context overwrites the get_context_data(), so it is somehow breaking the django’s action bean flow.

Question 5

So far I’ve only been able to access these url parameters from within the get_queryset method, although I’ve only tried it with a ListView not a TemplateView. I’ll use the url param to create an attribute on the object instance, then use that attribute in get_context_data to populate the context:

class Yearly(TemplateView):
    template_name = "calendars/yearly.html"

    current_year = datetime.datetime.now().year
    current_month = datetime.datetime.now().month

    def get_queryset(self):
        self.year = self.kwargs['year']
        queryset = super(Yearly, self).get_queryset()
        return queryset

    def get_context_data(self, **kwargs):
        context = super(Yearly, self).get_context_data(**kwargs)
        context['current_year'] = self.current_year
        context['current_month'] = self.current_month
        context['year'] = self.year
        return context

Question 6

How about just use Python decorators to make this intelligible:

class Yearly(TemplateView):

    @property
    def year(self):
       return self.kwargs['year']

Question 7

I have the following code for serializing the queryset;

def render_to_response(self, context, **response_kwargs):

    return HttpResponse(json.simplejson.dumps(list(self.get_queryset())),
                        mimetype="application/json")

And following is my get_querset()

[{'product': <Product: hederello ()>, u'_id': u'9802', u'_source': {u'code': u'23981', u'facilities': [{u'facility': {u'name': {u'fr': u'G\xe9n\xe9ral', u'en': u'General'}, u'value': {u'fr': [u'bar', u'r\xe9ception ouverte 24h/24', u'chambres non-fumeurs', u'chambres familiales',.........]}]

Which I need to serialize. But it says not able to serialize the <Product: hederello ()>. Because list composed of both django objects and dicts. Any ideas ?

Question 8

simplejson and json don’t work with django objects well.

Django’s built-in serializers can only serialize querysets filled with django objects:

data = serializers.serialize('json', self.get_queryset())
return HttpResponse(data, content_type="application/json")

In your case, self.get_queryset() contains a mix of django objects and dicts inside.

One option is to get rid of model instances in the self.get_queryset() and replace them with dicts using model_to_dict:

from django.forms.models import model_to_dict

data = self.get_queryset()

for item in data:
   item['product'] = model_to_dict(item['product'])

return HttpResponse(json.simplejson.dumps(data), mimetype="application/json")

Hope that helps.

Question 9

The easiest way is to use a JsonResponse.

For a queryset, you should pass a list of the the values for that queryset, like so:

from django.http import JsonResponse

queryset = YourModel.objects.filter(some__filter="some value").values()
return JsonResponse({"models_to_return": list(queryset)})

Question 10

I found that this can be done rather simple using the “.values” method, which also gives named fields:

result_list = list(my_queryset.values('first_named_field', 'second_named_field'))
return HttpResponse(json.dumps(result_list))

“list” must be used to get data as iterable, since the “value queryset” type is only a dict if picked up as an iterable.

Documentation: https://docs.djangoproject.com/en/1.7/ref/models/querysets/#values

Question 11

From version 1.9 Easier and official way of getting json

from django.http import JsonResponse
from django.forms.models import model_to_dict


return JsonResponse(  model_to_dict(modelinstance) )

Question 12

Our js-programmer asked me to return the exact JSON format data instead of a json-encoded string to her.

Below is the solution.(This will return an object that can be used/viewed straightly in the browser)

import json
from xxx.models import alert
from django.core import serializers

def test(request):
    alert_list = alert.objects.all()

    tmpJson = serializers.serialize("json",alert_list)
    tmpObj = json.loads(tmpJson)

    return HttpResponse(json.dumps(tmpObj))

Question 13

First I added a to_dict method to my model ;

def to_dict(self):
    return {"name": self.woo, "title": self.foo}

Then I have this;

class DjangoJSONEncoder(JSONEncoder):

    def default(self, obj):
        if isinstance(obj, models.Model):
            return obj.to_dict()
        return JSONEncoder.default(self, obj)


dumps = curry(dumps, cls=DjangoJSONEncoder)

and at last use this class to serialize my queryset.

def render_to_response(self, context, **response_kwargs):
    return HttpResponse(dumps(self.get_queryset()))

This works quite well

Python 实用宝典

标签归档：django-class-based-views

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