I would like a user to send a GET request to my Django REST API:
127.0.0.1:8000/model/?radius=5&longitude=50&latitude=55.1214
with his longitude/latitude and radius, passed in parameters, and get the queryset using GeoDjango.
For example, currently I have:
class ModelViewSet(viewsets.ModelViewSet):
queryset = Model.objects.all()
And what I ideally want is:
class ModelViewSet(viewsets.ModelViewSet):
radius = request.data['radius']
location = Point(request.data['longitude'],request.data['latitude']
# filter results by distance using geodjango
queryset = Model.objects.filer(location__distance_lte=(location, D(m=distance))).distance(location).order_by('distance')
Now a couple of immediate errors:
1) request is not defined - should I use api_view, i.e. the function based view for this?
2) DRF page says that request.data is for POST, PUT and PATCH methods only. How can send parameters with GET?
You can override get_queryset method for that purpose. As for query string parameters, you are right, request.data holds POST data, you can get query string params through request.query_params
def get_queryset(self):
longitude = self.request.query_params.get('longitude')
latitude= self.request.query_params.get('latitude')
radius = self.request.query_params.get('radius')
location = Point(longitude, latitude)
queryset = Model.objects.filter(location__distance_lte=(location, D(m=distance))).distance(location).order_by('distance')
return queryset
I had the same problem, to solve it you can get parameters from url with self.request.parser_context.get('kwargs') under the get_queryset method.
this actually worked for me .
using the self.request.query_params.get("lead_contact_id")
def get_queryset(self,*args,**kwargs):
# the lead id
lead_contact_id = self.request.query_params.get("lead_contact_id")
# this filter base on the lead id provided
feedback = Feedback.objects.filter(object_id=lead_contact_id)
return feedback
Related
I have a Class Based View to list animals from a specific herd. There are multiple herds, so the user can either see all animals from ONE herd, or all animals from ALL herds.
How do I have an optional URL parameter and handle it in the CBV?
urls:
url(r'list/(?P<hpk>[0-9]+)/$', AnimalList.as_view(), name = 'animal_list'),
url(r'list/$', AnimalList.as_view(), name = 'animal_list'),
My view:
class AnimalList(ListView):
model = Animal
def get_queryset(self):
if self.kwargs is None:
return Animal.objects.all()
return Animal.objects.filter(herd = self.kwargs['hpk']) # <--- line 19 that returns an error
Going to a URL of like /animals/list/3/ works fine, while /animals/list/ fails with an error. Here's that error:
KeyError at /animals/list/
'hpk'
Request Method: GET
Request URL: http://localhost:8000/animals/list/
Django Version: 1.8.2
Exception Type: KeyError
Exception Value:
'hpk'
Exception Location: /var/www/registry/animals/views.py in get_queryset, line 19
I get that the self.kwargs is a dictionary, and when I print() it inside the view, it'll show it's empty. But I can't figure out how to capture that scenario. I feel like this is a simple, stupid error I'm missing.
To anyone who may stumble on this and need an answer, here is my working code after figuring it out:
class AnimalList(ListView):
model = Animal
def get_queryset(self):
if 'hpk' in self.kwargs:
return Animal.objects.filter(herd = self.kwargs['hpk'])
return Animal.objects.all()
Essentially we test to see if the URL parameter hpk is present in the list of self.kwargs. If it is, we filter the queryset. Otherwise, we return all animals.
Hope this helps someone :)
I would implement this using GET parameters instead of separate URLs. With this approach, there is only one URL /list/ that is filtered by parameters, for example /list/?hpk=1.
This is more flexible as you can eventually add more queries /list/?hpk=1&origin=europe
#url(r'list/$', AnimalList.as_view(), name = 'animal_list'),
class AnimalList(ListView):
model = Animal
def get_queryset(self):
queryset = Animal.objects.all()
hpk = self.request.GET.get("hpk"):
if hpk:
try:
queryset = queryset.filter(herd=hpk)
except:
# Display error message
return queryset
The URL pattern having issues is:
url(r'^$', business_list, name='business_list_home'),
url(r'^(?P<param>\w+)$', business_list, name='business_list_results'),
My view is:
#render_to('app/business_list.html')
def business_list(request, param=None):
queryset = Business.objects.all()
search_form = SearchForm
print request.GET
if param in request.GET:
param = request.GET.get('param')
if queryset.filter(city__iexact=param).exists():
queryset = queryset.filter(city__iexact=param)
elif queryset.filter(category__iexact=param).exists():
queryset = queryset.filter(category__iexact=param)
print queryset
else:
queryset = None
print queryset
return {'business_list': queryset, 'search_form': search_form}
Essentially, I don't understand why when I go to /Miami I don't have access to it via request.GET['param'] in the view? print request.GET prints <QueryDict: {}>
The reason I want to do this is to have a nice URL scheme for displaying results of businesses for the city or category (and that's why I check if it's a city or category in the view too) in the url. Let me know if there's a better way to accomplish this like
url(r'^$', business_list),
url(r'^(?P<city>\w+)$', business_list),
url(r'^(?P<category>\w+)$', business_list),
url(r'^(?P<category>\w+)/(?P<city>\w+)$', business_list),
Thanks for any help!
business_list is a function that takes a request and a parameter called "param" (should probably give it a better name as to make it a little less generic). In Django, the URL routes can define what values get passed to the parameters of the view functions.
In the situation above, when a user goes to /Miami, Django will try to match the string "Miami" with any of the regular expressions defined in the URL routes. In this case, the matching expression is ^(?P<param>\w+)$. When this match is made, the string Miami is captured into param.
Django will then call business_list(request, param="Miami"). Note that there were no query parameters passed in the URL (e.g., /Miami?color=blue).
The issue in the code you've written above is that you're checking to see not if param exists, but rather that param was passed in the query parameters. Try going to the URL /Miami?param=test and it'll probably work the way that you expected.
The real fix here is to not reference request.GET, because using GET parameters is exactly what you're trying to avoid. So, instead of
if param in request.GET:
param = request.GET.get('param')
if queryset.filter(city__iexact=param).exists():
queryset = queryset.filter(city__iexact=param)
elif queryset.filter(category__iexact=param).exists():
queryset = queryset.filter(category__iexact=param)
print queryset
Just do:
if param:
if queryset.filter(city__iexact=param).exists():
queryset = queryset.filter(city__iexact=param)
elif queryset.filter(category__iexact=param).exists():
queryset = queryset.filter(category__iexact=param)
Full Disclosure: Cross posted to Tastypie Google Group
I have a situation where I have limited control over what is being sent to my api. Essentially there are two webservices that I need to be able to accept POST data from. Both use plain POST actions with urlencoded data (basic form submission essentially).
Thinking about it in "curl" terms it's like:
curl --data "id=1&foo=2" http://path/to/api
My problem is that I can't update records using POST. So I need to adjust the model resource (I believe) such that if an ID is specified, the POST acts as a PUT instead of a POST.
api.py
class urlencodeSerializer(Serializer):
formats = ['json', 'jsonp', 'xml', 'yaml', 'html', 'plist', 'urlencoded']
content_types = {
'json': 'application/json',
'jsonp': 'text/javascript',
'xml': 'application/xml',
'yaml': 'text/yaml',
'html': 'text/html',
'plist': 'application/x-plist',
'urlencoded': 'application/x-www-form-urlencoded',
}
# cheating
def to_urlencoded(self,content):
pass
# this comes from an old patch on github, it was never implemented
def from_urlencoded(self, data,options=None):
""" handles basic formencoded url posts """
qs = dict((k, v if len(v)>1 else v[0] )
for k, v in urlparse.parse_qs(data).iteritems())
return qs
class FooResource(ModelResource):
class Meta:
queryset = Foo.objects.all() # "id" = models.AutoField(primary_key=True)
resource_name = 'foo'
authorization = Authorization() # only temporary, I know.
serializer = urlencodeSerializer()
urls.py
foo_resource = FooResource
...
url(r'^api/',include(foo_resource.urls)),
)
In #tastypie on Freenode, Ghost[], suggested that I overwrite post_list() by creating a function in the model resource like so, however, I have not been successful in using this as yet.
def post_list(self, request, **kwargs):
if request.POST.get('id'):
return self.put_detail(request,**kwargs)
else:
return super(YourResource, self).post_list(request,**kwargs)
Unfortunately this method isn't working for me. I'm hoping the larger community could provide some guidance or a solution for this problem.
Note: I cannot overwrite the headers that come from the client (as per: http://django-tastypie.readthedocs.org/en/latest/resources.html#using-put-delete-patch-in-unsupported-places)
I had a similar problem on user creation where I wasn't able to check if the record already existed. I ended up creating a custom validation method which validated if the user didn't exist in which case post would work fine. If the user did exist I updated the record from the validation method. The api still returns a 400 response but the record is updated. It feels a bit hacky but...
from tastypie.validation import Validation
class MyValidation(Validation):
def is_valid(self, bundle, request=None):
errors = {}
#if this dict is empty validation passes.
my_foo = foo.objects.filter(id=1)
if not len(my_foo) == 0: #if object exists
foo[0].foo = 'bar' #so existing object updated
errors['status'] = 'object updated' #this will be returned in the api response
return errors
#so errors is empty if object does not exist and validation passes. Otherwise object
#updated and response notifies you of this
class FooResource(ModelResource):
class Meta:
queryset = Foo.objects.all() # "id" = models.AutoField(primary_key=True)
validation = MyValidation()
With Cathal's recommendation I was able to utilize a validation function to update the records I needed. While this does not return a valid code... it works.
from tastypie.validation import Validation
import string # wrapping in int() doesn't work
class Validator(Validation):
def __init__(self,**kwargs):
pass
def is_valid(self,bundle,request=None):
if string.atoi(bundle.data['id']) in Foo.objects.values_list('id',flat=True):
# ... update code here
else:
return {}
Make sure you specify the validation = Validator() in the ModelResource meta.
I'm trying to setup Django Haystack to search based on some pretty urls. Here is my urlpatterns.
urlpatterns += patterns('',
url(r'^search/$', SearchView(),
name='search_all',
),
url(r'^search/(?P<category>\w+)/$', CategorySearchView(
form_class=SearchForm,
),
name='search_category',
),
)
My custom SearchView class looks like this:
class CategorySearchView(SearchView):
def __name__(self):
return "CategorySearchView"
def __call__(self, request, category):
self.category = category
return super(CategorySearchView, self).__call__(request)
def build_form(self, form_kwargs=None):
data = None
kwargs = {
'load_all': self.load_all,
}
if form_kwargs:
kwargs.update(form_kwargs)
if len(self.request.GET):
data = self.request.GET
kwargs['searchqueryset'] = SearchQuerySet().models(self.category)
return self.form_class(data, **kwargs)
I keep getting this error running the Django dev web server if I try and visit /search/Vendor/q=Microsoft
UserWarning: The model u'Vendor' is not registered for search.
warnings.warn('The model %r is not registered for search.' % model)
And this on my page
The model being added to the query must derive from Model.
If I visit /search/q=Microsoft, it works fine. Is there another way to accomplish this?
Thanks for any pointers
-Jay
There are a couple of things going on here. In your __call__ method you're assigning a category based on a string in the URL. In this error:
UserWarning: The model u'Vendor' is not registered for search
Note the unicode string. If you got an error like The model <class 'mymodel.Model'> is not registered for search then you'd know that you haven't properly created an index for that model. However this is a string, not a model! The models method on the SearchQuerySet class requires a class instance, not a string.
The first thing you could do is use that string to look up a model by content type. This is probably not a good idea! Even if you don't have models indexed which you'd like to keep away from prying eyes, you could at least generate some unnecessary errors.
Better to use a lookup in your view to route the query to the correct model index, using conditionals or perhaps a dictionary. In your __call__ method:
self.category = category.lower()
And if you have several models:
my_querysets = {
'model1': SearchQuerySet().models(Model1),
'model2': SearchQuerySet().models(Model2),
'model3': SearchQuerySet().models(Model3),
}
# Default queryset then searches everything
kwargs['searchqueryset'] = my_querysets.get(self.category, SearchQuerySet())
I am writing an application in Django, which uses [year]/[month]/[title-text] in the url to identitfy news items. To manage the items I have defined a number of urls, each starting with the above prefix.
urlpatterns = patterns('msite.views',
(r'^(?P<year>[\d]{4})/(?P<month>[\d]{1,2})/(?P<slug>[\w]+)/edit/$', 'edit'),
(r'^(?P<year>[\d]{4})/(?P<month>[\d]{1,2})/(?P<slug>[\w]+)/$', 'show'),
(r'^(?P<year>[\d]{4})/(?P<month>[\d]{1,2})/(?P<slug>[\w]+)/save$', 'save'),
)
I was wondering, if there is a mechanism in Django, which allows me to preprocess a given request to the views edit, show and save. It could parse the parameters e.g. year=2010, month=11, slug='this-is-a-title' and extract a model object out of them.
The benefit would be, that I could define my views as
def show(news_item):
'''does some stuff with the news item, doesn't have to care
about how to extract the item from request data'''
...
instead of
def show(year, month, slug):
'''extract the model instance manually inside this method'''
...
What is the Django way of solving this?
Or in a more generic way, is there some mechanism to implement request filters / preprocessors such as in JavaEE and Ruby on Rails?
You need date based generic views and create/update/delete generic views maybe?
One way of doing this is to write a custom decorator. I tested this in one of my projects and it worked.
First, a custom decorator. This one will have to accept other arguments beside the function, so we declare another decorator to make it so.
decorator_with_arguments = lambda decorator: lambda * args, **kwargs: lambda func: decorator(func, *args, **kwargs)
Now the actual decorator:
#decorator_with_arguments
def parse_args_and_create_instance(function, klass, attr_names):
def _function(request, *args, **kwargs):
model_attributes_and_values = dict()
for name in attr_names:
value = kwargs.get(name, None)
if value: model_attributes_and_values[name] = value
model_instance = klass.objects.get(**model_attributes_and_values)
return function(model_instance)
return _function
This decorator expects two additional arguments besides the function it is decorating. These are respectively the model class for which the instance is to be prepared and injected and the names of the attributes to be used to prepare the instance. In this case the decorator uses the attributes to get the instance from the database.
And now, a "generic" view making use of a show function.
def show(model_instance):
return HttpResponse(model_instance.some_attribute)
show_order = parse_args_and_create_instance(Order, ['order_id'])(show)
And another:
show_customer = parse_args_and_create_instance(Customer, ['id'])(show)
In order for this to work the URL configuration parameters must contain the same key words as the attributes. Of course you can customize this by tweaking the decorator.
# urls.py
...
url(r'^order/(?P<order_id>\d+)/$', 'show_order', {}, name = 'show_order'),
url(r'^customer/(?P<id>\d+)/$', 'show_customer', {}, name = 'show_customer'),
...
Update
As #rebus correctly pointed out you also need to investigate Django's generic views.
Django is python after all, so you can easily do this:
def get_item(*args, **kwargs):
year = kwargs['year']
month = kwargs['month']
slug = kwargs['slug']
# return item based on year, month, slug...
def show(request, *args, **kwargs):
item = get_item(request, *args, **kwargs)
# rest of your logic using item
# return HttpResponse...