CodeJunkie
CodeJunkie

Reputation: 147

POST urls in tastypie

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

Answers (1)

sundar nataraj
sundar nataraj

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

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