Reputation: 147
i'm new to tastypie in django. I have an organisation model. In api.py
class OrganisationResource(ModelResource):
create_user = fields.ForeignKey(PersonResource, 'create_user', null=True, full=True)
update_user = fields.ForeignKey(PersonResource, 'update_user', null=True, full=True)
location = fields.ForeignKey(LocationResource, 'location', null=True, full=True)
class Meta:
allowed_methods = ['post','get','delete','patch','put']
queryset = Organization.objects.all()
resource_name = 'organisation'
authorization = Authorization()
authentication = Authentication()
always_return_data = True
The api url is,
http://127.0.0.1:8000/api/v1/organisation/
A post request to the above link is saving that data to database. But my doubt is can i use a seperate link to post or overwrite the current url so that i can send the post request to that link . Like
http://127.0.0.1:8000/api/v1/organisation/create
Upvotes: 1
Views: 182
Reputation: 8692
You can create seperate url for the Post using prependurls
useful link
class OrganisationCreateResource(ModelResource):
create_user = fields.ForeignKey(PersonResource, 'create_user', null=True, full=True)
update_user = fields.ForeignKey(PersonResource, 'update_user', null=True, full=True)
location = fields.ForeignKey(LocationResource, 'location', null=True, full=True)
class Meta:
allowed_methods = ['post']
queryset = Organization.objects.all()
detail_uri_name = 'create'
resource_name = 'organisation'
authorization = Authorization()
authentication = Authentication()
always_return_data = True
def prepend_urls(self):
return [
url(r'^(?P<resource_name>%s)/create/' % self._meta.resource_name, self.wrap_view('dispatch_detail'), name='api_dispatch_detail'),
]
Upvotes: 1