问Django模型表单同时保存两次post请求
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Stack Overflow用户

提问于 2019-01-04 15:17:56

回答 2查看 950关注 0票数 4

我在https://pypi.org/project/django-bootstrap-modal-forms/之后使用django-bootstrap-modal-forms 1.3.1，但如果我运行它的图书项目，它会调用post请求创建图书两次，但保存一次。

当我使用它时，它发出了两次post请求，但两个请求都保存了一个空文件，即一个post保存了像title description，date这样的模式的所有字段，而不是upload (文件)，而下一个post同时保存了所有的upload

这是我的模型：

class File(models.Model):
    id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False)
    author = models.ForeignKey(User, on_delete=models.CASCADE)
    visible_to_home = models.ManyToManyField(Home, blank=True)  # when none visible to all home
    visible_to_company = models.ManyToManyField(Company, blank=True)  # when none visible to all company
# To determine visibility, check if vtc is none or include company of user and if true, check same for home
    created_date = models.DateTimeField(auto_now=True)
    published = models.BooleanField(default=True)
    upload = models.FileField(blank=True, null=True, upload_to=update_filename)
    title = models.CharField(max_length=225, blank=True, null=True)
    description = models.TextField(blank=True, null=True)

如果我删除作者，那么它工作得很好，但作为我的要求，我需要作者。

class FileForm(PopRequestMixin, CreateUpdateAjaxMixin, forms.ModelForm):
    class Meta:
        model = File
        fields = ('title', 'description', 'upload')

视图

class FileCreateView(PassRequestMixin, SuccessMessageMixin,
                 CreateView):
    template_name = 'file/upload-file.html'
    form_class = FileForm
    success_message = 'File was uploaded successfully'
    success_url = reverse_lazy('home')

    def post(self, *args, **kwargs):
        """
        Handle POST requests: instantiate a form instance with the passed
        POST variables and then check if it's valid.
        """
        form = self.get_form()
        # form = self.form_class(self.request.POST, self.request.FILES)
        if self.request.method == 'POST':
            if form.is_valid():
                file = form.save(commit=False)
                file.upload = form.cleaned_data['upload']
                file.author = User.objects.get(pk=self.request.user.pk)
                file.save()
                return self.form_valid(form)
            else:
                return self.form_invalid(form)

home.html

{% block extrascripts %}
<script type="text/javascript">
$(function () {
  $(".upload-file").modalForm({formURL: "{% url 'file-upload' %}"});
 });
</script>
{% endblock extrascripts %}

Other与示例实现相同。

django

django-models

django-forms

回答 2

Stack Overflow用户

发布于 2020-03-25 19:57:32

试试这个：

 if form.is_valid():
     if not self.request.is_ajax():
         file = form.save(commit=False)
         file.upload = form.cleaned_data['upload']
         file.author = User.objects.get(pk=self.request.user.pk)
         file.save()

票数 3

Stack Overflow用户

发布于 2019-01-04 15:38:34

您需要修改您的CreateView

class FileCreateView(PassRequestMixin, SuccessMessageMixin,
                 CreateView):
    template_name = 'file/upload-file.html'
    form_class = FileForm
    success_message = 'File was uploaded successfully'
    success_url = reverse_lazy('home')

    def form_valid(self, form):
       file = form.save(commit=False)
       file.author = self.request.user.pk
       file.save()
       return HttpResponseRedirect(reverse('home'))