django-rest-framework:列出URL中的参数

Question

我是django和django-rest-framework的新手,但我试图将列表传递给url参数,然后通过它们过滤我的模型.

让我们说客户端应用程序发送的请求看起来像这样......

  url: "api.com/?something=string,string2,string3&?subthings=sub,sub2,sub3&?year=2014,2015,2016/"

我想用它们的值传递那些参数"东西","子东西"和"年".

网址看起来像这样？

注意:Trick是每个参数不会总是长度为3的数组.

有人能指出我正确的方向,我的url正则表达式应该如何处理列表,并在我的视图中检索查询列表.

谢谢!

Answer 1

通过上面的文档链接显示我是如何做到这一点的.注意:我使用管道作为我的网址分隔符而不是逗号 - >'|'.

在我的 urls.py

url(r'^$', SomethingAPIView.as_view(), name='something'),

在我的 views.py

class SomethingAPIView(ListAPIView):
  # whatever serializer class

  def get_queryset(self):
    query_params = self.request.query_params
    somethings = query_params.get('something', None)
    subthings = query_params.get('subthing', None)
    years = query_params.get('year', None)

    # create an empty list for parameters to be filters by 
    somethingParams = []
    subthingsParams = []
    yearParams = []

    # create the list based on the query parameters
    if somethings is not None:
      for something in somethings.split('|'):
        countryParams.append(int(something))
    if subthings is not None:
      for subthing in subthings.split('|'):
        subthingsParams.append(int(subthing))
    if years is not None:
      for year in years.split('|'):
        yearParams.append(int(year))

    if somethings and subthings and years is not None:
      queryset_list = Model.objects.all()
      queryset_list = queryset_list.filter(something_id__in=countryParams)
      queryset_list = queryset_list.filter(subthing_id__in=subthingsParams)
      queryset_list = queryset_list.filter(year__in=yearParams)
      return queryset_list

如果它们无效,我确实需要检查空结果.但是,对于希望在查询参数中传递多个值的人来说,这是起点.

这里有一个有效的网址/?something=1|2|3&subthing=4|5|6&year=2015|2016.

Answer 2

查看此文档http://www.django-rest-framework.org/api-guide/filtering/

查询参数通常不由url regex验证