TypeError:python中格式字符串的参数不足

Question

它显示了一个类似TypeError的错误：没有足够的参数来格式化字符串，如何解决这个问题。你知道吗

你知道吗视图.py你知道吗

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)

你知道吗网址.py你知道吗

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

Answer 1

在查询集中您有多个%s，尽管您通过了%(user_id)一次

您应该以您定义的格式字符串传递所有参数

cursor.execute("..." % (param1, param2, param3, param4,...))