Question

我希望处理我在django中使用HTML创建的表单，但我找不到从输入中获取值的正确方法。

这是我的HTML表单：

<form action="" class="form-inline" method="post">
{% csrf_token %}
    <!-- malfunction description -->
    <label class="requiredField" for="malfunctionDescription">תאור
        התקלה</label>
    <br/>
    <textarea class="form-control" style="width: 100%; margin-bottom: 3px"
              id="malfunctionDescription"
              name="malfunctionDescription"
              rows="5">
     </textarea>
</form>

这是我的view.py，很遗憾是空的：

def index(request):
error = ''
if request.method == 'POST':
    status = request.POST['status']
    rank = request.POST['rank']
    opener = request.POST['MalfunctionOpener']
    handler = request.POST['malfunctionHandler']
    system = request.POST['system']
    unit = request.POST['unit']
    opening_date = request.POST['openingdate']
    closing_date = request.POST['closingdate']
    description = request.POST['malfunctionDescription']
    solution = request.POST['malfunctionSolution']
    summary = request.POST['malfunctionSummary']

    find_description = Malfunction.objects.filter(description=description)
    if find_description:
        error = 'This malfunction is already stored in the data base'
    else:
        Malfunction.objects.create(status=status, rank=rank, opener=opener, handler=handler, system=system,
                                   unit=unit, openingDate=opening_date, closingDate=closing_date, solution=solution,
                                   summary=summary, description=description)

else:
    error = 'Something went wrong'

return render(request, 'resources-management/home.html')

我的主要目标是从表单中获取此信息并创建一个新对象，将其推送到数据库。如果我理解正确创建这个对象，我需要做类似的事情：

Object_name.object.create(information=information)

但我不知道如何获取信息，我需要像烧瓶中的 request.form [＆＃39; name＆＃39;]

谢谢！

----编辑1 - url.py ----

这是主要的url.py

urlpatterns = [
url(r'^', include('elbit_ils.urls')),
url(r'^admin/', admin.site.urls),
url(r'^resources-management/', include('resources_management.urls')),
url(r'^registration-login/', include('registration_login.urls')),
url(r'^contact/', include('contact.urls')),

这是应用程序url.py

urlpatterns = [ url(r'^$', IndexView.as_view(), name="my_list"), url(r'^(?P<pk>\d+)$', DetailView.as_view(model=Malfunction, template_name="resources-management/malfunction.html")), ]

- 编辑2：indexview -

类IndexView（CreateView）：

context_object_name = 'my_list'
template_name = 'resources-management/home.html'
queryset = Malfunction.objects.all()
fields = '__all__'

def get_context_data(self, **kwargs):
    context = super(IndexView, self).get_context_data(**kwargs)
    context['contacts'] = Contact.objects.all().order_by("firstname")
    context['malfunctions'] = Malfunction.objects.all().order_by("-openingDate")
    context['systems'] = System.objects.all().order_by("systemname")
    context['units'] = Unit.objects.all().order_by("unitname")
    # And so on for more models
    return context

Answer 1

如果你有一个像

这样的表格

<form action="" method="post">
    <input type="text" name="username" />
    <input type="submit" />
</form>

请注意name元素中的input属性。然后

在django中，您将获得request.POST中的数据。所以您可以

request.POST['username']

Answer 2

您应该能够以下列方式访问数据

def index(request):
    # All the post data is stored in `request.POST`
    desc = request.POST["malfunctionDescription"]

您可以查看如何在文档中使用Forms，以了解验证所收到数据的最佳方法。

Django - 没有表单模型的处理表单

2 个答案: