Python get_model_by_resource_url Examples

Example #1

File: api.py Project: TrevorFSmith/transmutable

	def create_detail(self, object_list, bundle):
		if not bundle.request.user.is_authenticated(): return False
		if not 'namespace' in bundle.data: return False
		namespace = get_model_by_resource_url(bundle.data['namespace'], Namespace)
		if not namespace: return False
		if namespace.owner != bundle.request.user: return False
		return True

Example #2

 def create_detail(self, object_list, bundle):
     if not bundle.request.user.is_authenticated(): return False
     if not 'namespace' in bundle.data: return False
     namespace = get_model_by_resource_url(bundle.data['namespace'],
                                           Namespace)
     if not namespace: return False
     if namespace.owner != bundle.request.user: return False
     return True

Example #3

def get_namespace_by_resource_url(url):
    '''Returns a Namespace for a resource URL like /api/v0.1/peach/namespace/3/'''
    return get_model_by_resource_url(url, Namespace)

Example #4

File: api.py Project: TrevorFSmith/transmutable

def get_namespace_by_resource_url(url):
	'''Returns a Namespace for a resource URL like /api/v0.1/peach/namespace/3/'''
	return get_model_by_resource_url(url, Namespace)