TypeError:python中格式字符串的参数不足

它显示了一个类似TypeError的错误：格式字符串的参数不够，如何解决这个问题

views.py

def GetMobilefollowpopularnewsproviderTest(request,user_id):
# from django.http import JsonResponse
print "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
if request.method == 'GET':
    # news_provider = request.POST.get("news_provider")
    print user_id
    cursor = connection.cursor()
    queryset = cursor.execute("select news_crawl_newsproviders.news_provider as id, 1 as status from news_crawl_newsproviders join  accounts_follownewsprovider on news_crawl_newsproviders.id=accounts_follownewsprovider.provider_id where accounts_follownewsprovider.user_id='%s' union select news_crawl_newsproviders.news_provider, 0 from news_crawl_newsproviders join  accounts_follownewsprovider on news_crawl_newsproviders.id=accounts_follownewsprovider.provider_id where accounts_follownewsprovider.user_id!='%s' and news_crawl_newsproviders.news_provider not in(select news_crawl_newsproviders.news_provider as id from news_crawl_newsproviders join  accounts_follownewsprovider on news_crawl_newsproviders.id=accounts_follownewsprovider.provider_id where accounts_follownewsprovider.user_id='%s');select news_crawl_newsproviders.news_provider as id, 1 as status from news_crawl_newsproviders join  accounts_follownewsprovider on news_crawl_newsproviders.id=accounts_follownewsprovider.provider_id where accounts_follownewsprovider.user_id='%s' union select news_crawl_newsproviders.news_provider, 0 from news_crawl_newsproviders join  accounts_follownewsprovider on news_crawl_newsproviders.id=accounts_follownewsprovider.provider_id where accounts_follownewsprovider.user_id!='%s' and news_crawl_newsproviders.news_provider not in(select news_crawl_newsproviders.news_provider as id from news_crawl_newsproviders join  accounts_follownewsprovider on news_crawl_newsproviders.id=accounts_follownewsprovider.provider_id where accounts_follownewsprovider.user_id='%s');"%(user_id))
    dict = {}
    dict = dictfetchall(cursor)
    print(dict)
    context = {
        'posts': dict
    }
    return JsonResponse(context, safe=False)
return HttpResponse(status=201)

url.py

url(r'followpopularnewsprovider/(?P<user_id>\d+)/$', 
 csrf_exempt(views.GetMobilefollowpopularnewsproviderTest), name='popularprovider')

url（r'followPopularNewProvider/（？P\d+）/$），
csrf_豁免（views.GetMobilefollowpopularnewsproviderTest），name='popularprovider'）

在您的查询集中您有多个

%s

，尽管您通过了

%（用户id）

您应该以定义的格式字符串传递所有参数

cursor.execute（“…”%（param1，param2，param3，param4，…）

在您的查询集中，您有两个

%s

虽然您通过了

%（用户id）

一次，但如何通过多个用户id查看下面的答案