我在Django創建了一個形式如下:創建在Django形式不是基於模型會導致錯誤
class advancedsearch(forms.Form):
choices = forms.ChoiceField (label="choices", choices = ("1", "2", "3"), required= False)
現在我想要在視圖中使用如下:
def advancedsearch(request):
# if request.method == "POST":
# search = advancedsearch(request.POST, request.FILES)
# if search.is_valid():
# return HttpResponseRedirect(reverse("view.designs")) # UPDATE REDIRECT
# else:
# print "FORM IS NOT VALID"
# # GET request
# else:
# search = advancedsearch()
# return render_to_response("advancedsearch.html", {
# "search": search,
# }, context_instance=RequestContext(request,{}))
# # generic case
search = advancedsearch()
return render_to_response("advancedsearch.html", {
"search": search,
}, context_instance=RequestContext(request, {}))
我得到的錯誤是advancedsearch() takes exactly 1 argument (0 given)
我不認爲我需要傳遞函數的參數 - 但它看起來像我...我應該通過它?
謝謝,但現在我得到的錯誤:「當我嘗試在我的模板中使用它作爲'{{search.choices}}'...時,需要多個值來解壓縮...」是否與此相關? – user1328021 2013-05-07 20:04:34
檢查編輯.. – karthikr 2013-05-07 20:05:27
謝謝!不知道我是如何錯過的... – user1328021 2013-05-07 20:07:10