我是django的新手。我有一个带有视图的新闻应用程序,它为每个新闻呈现一个页面:
def news_page(request, news_id):
news = News.objects.get(pk=news_id)
tags = news.tags.all()
category = news.category
comments = news.comment_set.all()
form = add_comment(request, news.id)
return render(request, 'news/news_page.html', {'form': form, 'news': news, 'tags': tags, 'category': category, 'user': request.user, 'comments': comments})
我已经创建了评论应用: models.py:
class Comment(models.Model):
author = models.ForeignKey(User)
comment_body = models.CharField(max_length=500)
news = models.ForeignKey(News)
pub_date = models.DateTineField(default = datetime.datetime.now())
forms.py:
class AddCommentForm(ModelForm):
comment_body = forms.CharField(widget=forms.Textarea)
class Meta():
model = Comment
exclude = ('author', 'news','pub_date',)
我试着实现comments.view function add_comment news_page view(上面)使用:
def add_comment(request, news_id):
news = News.objects.get(pk=news_id)
if request.method == 'POST':
form = AddCommentForm(request.POST)
if form.is_valid:
comment = form.save(commit=False)
comment.author = request.user
comment.news = news
comment.save()
else:
form = AddCommentForm()
return form
但我有错误'AddCommentForm'对象没有属性'has_header'。我认为这是因为add_comment视图没有HttpResponse。我应该如何重写代码以使这个想法变得有效。 错误追溯
Internal Server Error: /news/15/comment/
Traceback (most recent call last):
File "C:\Python27\myproject\djcode\first_venv\venv\lib\site-packages\django\core\handlers\base.py"
, line 187, in get_response
response = middleware_method(request, response)
File "C:\Python27\myproject\djcode\first_venv\venv\lib\site-packages\django\contrib\sessions\middl
eware.py", line 26, in process_response
patch_vary_headers(response, ('Cookie',))
File "C:\Python27\myproject\djcode\first_venv\venv\lib\site-packages\django\utils\cache.py", line
142, in patch_vary_headers
if response.has_header('Vary'):
AttributeError: 'AddCommentForm' object has no attribute 'has_header'
感谢的!
答案 0 :(得分:8)
从视图中你需要返回一个response
对象,但是你要返回表单对象。因此错误。
您可以使用render
代替
更改
return form
到
return render(request, template_name, {'form': form})
答案 1 :(得分:0)
karthikr是对的
您也可以使用
return render_to_response('template_name',{'form':form})