Django 如何将PK传递给dispatch_detail

Django 如何将PK传递给dispatch_detail,django,tastypie,Django,Tastypie,以下是我为资源列表提供的url: url(r'^(?P<resource_name>%s)/stats%s$' % ( self._meta.resource_name, trailing_slash()), self.wrap_view('dispatch_list'), name='api_dispatch_regions_stats') 通常你会这么做 url(r'^(?P<resource_name>

以下是我为资源列表提供的url:

url(r'^(?P<resource_name>%s)/stats%s$' % (
                self._meta.resource_name, trailing_slash()),
                self.wrap_view('dispatch_list'), name='api_dispatch_regions_stats')
通常你会这么做

url(r'^(?P<resource_name>%s)/(?P<pk>\d+)/stats%s$' % (self._meta.resource_name, trailing_slash()),
    self.wrap_view('dispatch_list'),
    name='api_dispatch_regions_stats')
url(r'^(?P%s)/(?P\d+)/stats%s$'%(self.\u meta.resource\u name,trailing\u slash()),
self.wrap_视图(“调度列表”),
name='api\u调度\u区域\u统计')
Django关心其余的事情,但我不确定你到底想做什么

def alter_detail_data_to_serialize(self, request, data):
    if 'stats' in request.path:
        do_something()
    return data
url(r'^(?P<resource_name>%s)/(?P<pk>\d+)/stats%s$' % (self._meta.resource_name, trailing_slash()),
    self.wrap_view('dispatch_list'),
    name='api_dispatch_regions_stats')