使用 django sweetpie 创建一个搜索 API

Question

我正在尝试设计一个用于搜索功能的 api。我想要一个在 reactjs 中实现的 API。我想要的是 /api/v1/rent/search/place="place name" 但我没有得到这个。我所做的是

api.py

from rentals.models import Rental,Gallery
from django.core.paginator import InvalidPage
from django.conf.urls import *
from tastypie.paginator import Paginator
from tastypie.exceptions import BadRequest
from tastypie.resources import ModelResource
from tastypie.utils import trailing_slash
from haystack.query import SearchQuerySet

class SearchResource(ModelResource):
    class Meta:
        queryset = Rental.objects.all()
        resource_name = 'rent'

    def prepend_urls(self):
        return [
            url(r"^(?P<resource_name>%s)/search%s$" % (
                    self._meta.resource_name,
                    trailing_slash()),
                self.wrap_view('get_search'),
                name="api_get_search"
            ),
        ]

    def get_search(self, request, **kwargs):
        self.method_check(request, allowed=['get'])
        self.is_authenticated(request)
        self.throttle_check(request)

        # Do the query.
        sqs = SearchQuerySet().models(Rental).load_all().auto_query(request.GET.get('q', ''))
        paginator = Paginator(sqs, 20)

        try:
            page = paginator.page(int(request.GET.get('page', 1)))
        except InvalidPage:
            raise Http404("Sorry, no results on that page.")

        objects = []

        for result in page.object_list:
            bundle = self.build_bundle(obj=result.object, request=request)
            bundle = self.full_dehydrate(bundle)
            objects.append(bundle)

        object_list = {
            'objects': objects,
        }

        self.log_throttled_access(request)
        return self.create_response(request, object_list)

models.py

class Rental(models.Model):
    city =  models.CharField(_("City"), max_length=255, blank=False,null=True,
        help_text=_("City of the rental space"))
    place =  models.CharField(_("Place"), max_length=255, blank=False,null=True,
        help_text=_("Place of the rental space"))

class Gallery(models.Model):
    rental = models.ForeignKey('Rental', null=True, on_delete=models.CASCADE,verbose_name=_('Rental'), related_name="gallery")
    image = models.ImageField(blank=True,upload_to='upload/',null=True)

我还需要做什么才能实现 api/v1/rent/search/place="place name"（我想从地名搜索）这样的 url？

我收到以下错误

Answer 1

将此添加到 Python 文件的顶部：

from django.core.paginator import InvalidPage