I need to create an API that takes application/x-www-form-urlencoded data in POST data and returns a HTML content in response.
Now to serialize the urlencoded data I am calling following serializer from resource :
class urlencodeSerializer(Serializer):
formats = ['json', 'jsonp', 'xml', 'yaml', 'html', 'plist', 'urlencode']
content_types = {
'json': 'application/json',
'jsonp': 'text/javascript',
'xml': 'application/xml',
'yaml': 'text/yaml',
'html': 'text/html',
'plist': 'application/x-plist',
'urlencode': 'application/x-www-form-urlencoded',
}
def from_urlencode(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
def to_urlencode(self,content):
pass
To specify the response format I have added this function in the resource :
def determine_format(self, request):
return 'text/html'
Now as I am trying to output the HTML response as below :
data = "<html><h1>Hello</h1></html>"
return HttpResponse(data, content_type='text/html', status=200)
I am getting following error :
Sorry, not implemented yet. Please append "?format=json" to your URL.
Can anybody suggest me what is wrong with this code and how to achieve the given requirement.
I have found the answer though not sure if its the correct way to do in tastypie :
Normally return the bundle response from resource.
Add following method in urlencodeSerializer
def to_html(self,bundle, options):
return prepare_html_from_bundle_data(bundle.data)
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