Python 使用django tastypie创建搜索API

我正在尝试为搜索功能设计一个api。我想要一个用于在reactjs中实现的API。我想要的是/api/v1/rent/search/place=“地名”，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)

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)

要实现api/v1/rent/search/place=“place name”（我想从地名搜索）这样的url，我还需要做些什么

我得到以下错误

---

将以下内容添加到Python文件的顶部：

from django.core.paginator import InvalidPage

---

当我执行此api/v1/searchRent/search/？format=json时出现错误，您会遇到什么错误？如果我现在在问题中附加了错误，请提供整个stacktrace。如果我添加了name“/？format=json，则会出现错误。当我添加localhost:8000/api/v1/rent/？format=json时，请尝试localhost:8000/api/v1/rent/？format=json&q=“place%20name”“我不仅得到了名为california的place的结果，还得到了所有的place。Try:localhost:8000/api/v1/rent/search/？format=json&q=“california”“error\u message:“page（）接受1个位置参数，但给出了2个位置参数”，如果我这样做，将得到此错误。”