它显示了一个类似于TypeError的错误:格式字符串的参数不足,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)
urls . py
url(r'followpopularnewsprovider/(?P<user_id>\d+)/$',
csrf_exempt(views.GetMobilefollowpopularnewsproviderTest), name='popularprovider')
问题来源StackOverflow 地址:/questions/59386158/typeerror-not-enough-arguments-for-format-string-in-python
在您的queryset中,您有不止一个%s,尽管您一次传递了%(user_id) 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,...))
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