API を作成していたモバイル アプリケーション用に、同様の「検索」タイプのリソースを作成しました。基本的に、アプリケーションの検索フィードに表示する一連のタイプといくつかの共通フィールドについて合意しました。実装については、以下のコードを参照してください。
class SearchObject(object):
def __init__(self, id=None, name=None, type=None):
self.id = id
self.name = name
self.type = type
class SearchResource(Resource):
id = fields.CharField(attribute='id')
name = fields.CharField(attribute='name')
type = fields.CharField(attribute='type')
class Meta:
resource_name = 'search'
allowed_methods = ['get']
object_class = SearchObject
authorization = ReadOnlyAuthorization()
authentication = ApiKeyAuthentication()
object_name = "search"
include_resource_uri = False
def detail_uri_kwargs(self, bundle_or_obj):
kwargs = {}
if isinstance(bundle_or_obj, Bundle):
kwargs['pk'] = bundle_or_obj.obj.id
else:
kwargs['pk'] = bundle_or_obj['id']
return kwargs
def get_object_list(self, bundle, **kwargs):
query = bundle.request.GET.get('query', None)
if not query:
raise BadRequest("Missing query parameter")
#Should use haystack to get a score and make just one query
objects_one = ObjectOne.objects.filter(name__icontains=query).order_by('name').all)[:20]
objects_two = ObjectTwo.objects.filter(name__icontains=query).order_by('name').all)[:20]
objects_three = ObjectThree.objects.filter(name__icontains=query).order_by('name').all)[:20]
# Sort the merged list alphabetically and just return the top 20
return sorted(chain(objects_one, objects_two, objects_three), key=lambda instance: instance.identifier())[:20]
def obj_get_list(self, bundle, **kwargs):
return self.get_object_list(bundle, **kwargs)