Django：「對象不可迭代」，當試圖修改所述對象

Question

我想編輯一個已經創建的對象（「擔心」）。

我有這樣的代碼在views.py：

def worry_edit(request, id): 
    worry = get_object_or_404(Worry, pk = id) 
    original_pub_date = worry.pub_date 
    form = WorryForm(request.POST or None, instance = worry) 
    if form.is_valid(): 
     worry = form.save(commit = False) 
     worry.pub_date = original_pub_date 
     worry.save() 
     return redirect(worry) 
    return render_to_response('holaProy/worry_edit.html', {'worry_form': form, 'worry_id': id}, context_instance=RequestContext(request)) 

擊中形式發送鍵時提供了以下錯誤：

argument of type 'Worry' is not iterable 

上爲什麼錯誤出現和任何見解如何解決它？

編輯：由於Girasquid建議，這是完整的回溯：

Environment: 

Request Method: POST 
Request URL: http://127.0.0.1:8000/holaProy/worry_edit/1/ 

Django Version: 1.4.1 
Python Version: 2.7.3 
Installed Applications: 
('django.contrib.auth', 
'django.contrib.contenttypes', 
'django.contrib.sessions', 
'django.contrib.sites', 
'django.contrib.messages', 
'django.contrib.staticfiles', 
'django.contrib.admin', 
'django.contrib.admindocs', 
'holaProy', 
'registration', 
'django.contrib.humanize') 
Installed Middleware: 
('django.middleware.common.CommonMiddleware', 
'django.contrib.sessions.middleware.SessionMiddleware', 
'django.middleware.csrf.CsrfViewMiddleware', 
'django.contrib.auth.middleware.AuthenticationMiddleware', 
'django.contrib.messages.middleware.MessageMiddleware') 


Traceback: 
File "C:\Python27\lib\site-packages\django\core\handlers\base.py" in get_response 
    111.       response = callback(request, *callback_args, **callback_kwargs) 
File "C:\Users\Richard\proyectosPython\holaProyecto\holaProyecto\holaProy\views.py" in worry_edit 
    50.   return redirect(worry) 
File "C:\Python27\lib\site-packages\django\shortcuts\__init__.py" in redirect 
    81.   if '/' not in to and '.' not in to: 

Exception Type: TypeError at /holaProy/worry_edit/1/ 
Exception Value: argument of type 'Worry' is not iterable 

Answer 1

爲了能夠對模型實例傳遞給redirect，你需要在其上定義一個get_absolute_url()方法。否則，Django會認爲它是一個字符串形式的URL，並試圖解析它。