I'm using django tastypie to publish a model with a Related (ToOne) field to another model resource. The uri is:
/api/map/?format=json
I want to let the client include a full_pages url parameter to get the full related page resource: /api/map/?full_pages=1&format=json
I don't really understand the Relationship Fields docs, but I made a get_full callable:
def get_full(bundle):
if bundle.request.GET.get('full_pages', 0):
return True
return False
I tried passing the callable to the full argument of ToOneField:
from tastypie.contrib.gis import resources as gis_resources
class MapResource(gis_resources.ModelResource):
page = fields.ToOneField('pages.api.PageResource', 'page', full=get_full)
But when I check with pdb, get_full is never invoked.
So then I tried creating a custom FillableToOneField with a full attribute:
class FillableToOneField(fields.ToOneFIeld):
full = get_full
class MapResource(ModelResource):
page = FillableToOneField('pages.api.PageResource', 'page')
Again, get_full is never invoked.
Is there a better, easier way to do this?
After reading Amyth's answer and django-boundaryservice code, I got this to work by defaulting full to True and altering it in the dehydrate method on the Related PageResource:
class MapResource(gis_resources.ModelResource):
page = fields.ToOneField('pages.api.PageResource', 'page', full=True)
pages.api:
class PageResource(ModelResource):
...
def dehydrate(self, bundle):
if not bundle.request.GET.get('full_pages'):
bundle = bundle.data['resource_uri']
return bundle
You can simply achieve this under the dehydrate method as follows.
class MapResource(ModelResource):
page = fields.ToOneField('pages.api.PageResource', 'page')
def dehydrate(self, bundle):
if bundle.request.Get.get('full_pages'):
self.page.full = True
return bundle
and have them send a request as /api/map/?full_pages=True&format=json
Related
Following Django-Rest-Framework Doc about Custom metadata I need to add an attribute (myatt) to determine_metadata() method, like this for example:
def determine_metadata(self, request, view):
metadata = OrderedDict()
metadata['name'] = view.get_view_name()
metadata['description'] = view.get_view_description()
metadata['myatt'] = 'blablabla'
metadata['renders'] = [renderer.media_type for renderer in view.renderer_classes]
metadata['parses'] = [parser.media_type for parser in view.parser_classes]
if hasattr(view, 'get_serializer'):
actions = self.determine_actions(request, view)
if actions:
metadata['actions'] = actions
return metadata
I edited the metadata.py file in rest-framework directory just to test it and worked. However in the correct way, I know that I shall override the determine_metadata() method to accomplish what I want.
My problem is that I don't know where and how shall I override it.
Can you help me?
You can create a MyCustomMetadata class which will add an extra attribute to the metadata.
This class will inherit from SimpleMetadata class which is the default metadata class in DRF. Then, we will override the determine_metadata() function in it. We first call the super() to get the original metadata returned by DRF, then we add our extra attribute to the metadata.
my_app/metadata.py
from rest_framework.metadata import SimpleMetadata
class MyCustomMetadata(SimpleMetadata):
def determine_metadata(self, request, view):
metadata = super(MyCustomMetadata, self).determine_metadata(request, view)
metadata['myatt'] = 'blablabla' # add extra attribute to metadata
return metadata # return the metadata with the extra attribute set in it
Now, we need to define MyCustomMetadata class in our settings which will be used by DRF then.
settings.py
REST_FRAMEWORK = {
...
'DEFAULT_METADATA_CLASS': 'my_app.metadata.MyCustomMetadata'
}
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
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'm currently working with django generic views and I have a problem I can't figure out.
When using delete_object I get a TypeError exception:
delete_object() takes at least 3 non-keyword arguments (2 given)
Here is the code (I have ommited docstrings and imports):
views.py
def delete_issue(request, issue_id):
return delete_object(request,
model = Issue,
object_id = issue_id,
template_name = 'issues/delete.html',
template_object_name = 'issue')
urls.py
urlpatterns = patterns('issues.views',
(r'(?P<issue_id>\d+)/delete/$', 'delete_issue'),
)
The other generic views (object_list, create_object, etc.) work fine with those parameters. Another problem I have is when using the create_object() function, it says something about a CSRF mechanism, what is that?
You need to provide post_delete_redirect, this means url, where user should be redirected after object is deleted. You can find this in view signature:
def delete_object(request, model, post_delete_redirect, object_id=None,
slug=None, slug_field='slug', template_name=None,
template_loader=loader, extra_context=None, login_required=False,
context_processors=None, template_object_name='object'):