Question

我正在寻找这个常见问题的简单答案。到目前为止，我找到的答案为我们的初学者留下了关键点。

我有一个应用程序，几乎每个模型都有一个ForeignKey to User，并且有一个unique_together约束，其中一个字段始终是'user'。

例如：

class SubscriberList(models.Model):

user = models.ForeignKey(User)
name = models.CharField(max_length=70)
date_created = models.DateTimeField(auto_now_add=True)

class Meta:
    unique_together = (
        ('user', 'name',),
        )   

def __unicode__(self):
    return self.name

订阅者列表始终由登录用户创建，因此在创建订阅者列表的表单中，我在保存表单时排除用户字段并为其赋予值self.request.user，如下所示： / p>

class SubscriberListCreateView(AuthCreateView):
model = SubscriberList
template_name = "forms/app.html"
form_class = SubscriberListForm
success_url = "/app/lists/"

def form_valid(self, form):
    self.object = form.save(commit=False)
    self.object.user = self.request.user
    return super(SubscriberListCreateView, self).form_valid(form)

以下是随附的表格：

class SubscriberListForm(ModelForm):
class Meta:
    model = SubscriberList
    exclude = ('user')

使用此代码，有效数据很好。当我提交不 unique_together的数据时，我从数据库中收到完整性错误。原因很清楚--Django没有验证unique_together，因为'user'字段被排除在外。

如何更改现有代码，仍然使用CreateView，以便提交的非unique_together数据会引发表单验证错误，而不会引发数据库中的完整性错误。

Answer 1

Yehonatan的例子让我在那里，但我不得不从form_valid的ValidationError中调用消息，而不是单独的form_invalid函数。

这有效：

class SubscriberCreateView(AuthCreateView):
    model = Subscriber
    template_name = "forms/app.html"
    form_class = SubscriberForm
    success_url = "/app/subscribers/"

    def form_valid(self, form):
        self.object = form.save(commit=False)
        self.object.user = self.request.user

        try:
            self.object.full_clean()
        except ValidationError:
            #raise ValidationError("No can do, you have used this name before!")
            #return self.form_invalid(form)
            from django.forms.util import ErrorList
            form._errors["email"] = ErrorList([u"You already have an email with that name man."])
            return super(SubscriberCreateView, self).form_invalid(form)

        return super(SubscriberCreateView, self).form_valid(form)

Answer 2

从以下文档中获取： https://docs.djangoproject.com/en/dev/ref/models/instances/?from=olddocs#validating-objects

如果您打算自己处理验证错误，或者您已从ModelForm中排除了需要验证的字段，则只需要调用模型的full_clean（）方法。

从以下文档中获取： https://docs.djangoproject.com/en/dev/ref/class-based-views/#formmixin

混合FormMixin的视图必须提供form_valid（）和form_invalid（）的实现。

这意味着为了查看错误（与表单无关），您需要实现自己的form_invalid，在那里添加特殊错误消息并将其返回。

因此，在对象上运行full_clean（）会引发unique_together错误，因此您的代码可能如下所示：

def form_valid(self, form):
    self.object = form.save(commit=False)
    self.object.user = self.request.user
    # validate unique_together constraint
    try:
        self.object.full_clean()
    except ValidationError:
        # here you can return the same view with error messages
        # e.g.
        return self.form_invalid(form)
    return super(SubscriberListCreateView, self).form_valid(form)

def form_invalid(self, form):
    # using messages
    # from django.contrib import messages
    # messages.error('You already have a list with that name')
    # or adding a custom error
    from django.forms.util import ErrorList
    form._errors["name"] = ErrorList([u"You already have a list with that name"])
    return super(SubscriberListCreateView, self).form_invalid(form)

HTH

Answer 3

添加另一个可能对新手更容易一些的例子。

forms.py

class GroupItemForm(ModelForm):
    def form_valid(self):
        self.object = self.save(commit=False)
        try:
            self.object.full_clean()
        except ValidationError:
            # here you can return the same view with error messages
            # e.g. field level error or...
            self._errors["sku"] = self.error_class([u"You already have an email with that name."])
            # ... form level error
            self.errors['__all__'] = self.error_class(["error msg"]
            return False
        return True

views.py

def add_stock_item_detail(request, item_id, form_class=GroupItemForm, template_name="myapp/mytemplate.html"):
    item = get_object_or_404(Item, pk=item_id)
    product = Product(item=item)
    if request.method == 'POST':
        form = form_class(request.POST, instance=product)
        if form.is_valid() and form.form_valid():
            form.save()
            return HttpResponseRedirect('someurl')
    else:
        form = form_class(instance=product)

    ctx.update({
        "form" : form,
     })

    return render_to_response(template_name, RequestContext(request, ctx))

Django 1.3 CreateView / ModelForm：unique_together验证，从表单中排除一个字段

3 个答案: