Convert POST to PUT with Tastypie - django

Question

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)

Answer 1

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...

Answer 2

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()

Answer 3

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.