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Python Django Generic View form_valid无法在指定中间模型的ManyToManyField上设置值_Python_Django_Many To Many - Fatal编程技术网

Python Django Generic View form_valid无法在指定中间模型的ManyToManyField上设置值

Python Django Generic View form_valid无法在指定中间模型的ManyToManyField上设置值,python,django,many-to-many,Python,Django,Many To Many,我有三个多对多关系表,一个产品表、一个服务器表和一个中介表,用于将两个对象链接在一起 每个服务器可以有许多产品,并且每个产品可以与多个服务器关联 这是我的模型 #/myapp/models.py class Server(TimeStampedModel): name = models.CharField(max_length=35) description = models.CharField(max_length=200) products = models.Man

我有三个多对多关系表,一个产品表、一个服务器表和一个中介表,用于将两个对象链接在一起

每个服务器可以有许多产品,并且每个产品可以与多个服务器关联

这是我的模型

#/myapp/models.py

class Server(TimeStampedModel):
    name = models.CharField(max_length=35)
    description = models.CharField(max_length=200)
    products = models.ManyToManyField('Product', through='ServerProduct',
                                      related_name='products')

class ServerProduct(TimeStampedModel):
    server = models.ForeignKey('Server', on_delete=models.CASCADE)
    product = models.ForeignKey('Product', on_delete=models.CASCADE)


class Product(TimeStampedModel):
    name = models.CharField(max_length=200)
    price = models.DecimalField(decimal_places=2, max_digits=11)
    servers = models.ManyToManyField(
        'Server', through='ServerProduct', related_name='servers')
在我的创建视图中,我指向一个表单,允许用户创建一个服务器,并选择它对应的产品

在form_valid()中,我试图将每个产品链接到新服务器

#/myapp/views.py

class ServerCreateView(SuccessMessageMixin, CreateView):
        model = Server
        form_class = ServerForm
        ....

        def form_valid(self, form):
            server = form.save(False)
            server.save()
            for product in form.cleaned_data['products']:
                ServerProduct.objects.create(server=server, product=product)
            return super(ServerCreateView, self).form_valid(form)
我的表格如下

class ServerForm(BlankToRequiredMixin):

    class Meta:
        model = Server
        fields = '__all__'
        widgets = {
            'name': forms.TextInput(attrs={'autofocus': 'autofocus'}),
        }
但是,当我提交表单django时,返回以下错误:

无法在指定中介的ManyToManyField上设置值 模型改用reports.ServerProduct的管理器

在阅读文档后,我也尝试了以下操作,但返回了相同的错误,以代替
ServerProduct.objects.create(server=server,product=product)

prod = ServerProduct(server=server, product=product)
prod.save()
你知道我该怎么解决这个问题吗?(最好仍然使用通用创建视图)

编辑:完全回溯

Environment:


Request Method: POST
Request URL: http://localhost:8000/server-create/

Django Version: 1.9.7
Python Version: 3.4.2
Installed Applications:
['django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'django_extensions',
 'auditlog',
 'rest_framework',
 'reports.apps.ReportsConfig']
Installed Middleware:
['django.middleware.security.SecurityMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.auth.middleware.SessionAuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware',
 'auditlog.middleware.AuditlogMiddleware']



Traceback:

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/core/handlers/base.py" in get_response
  149.                     response = self.process_exception_by_middleware(e, request)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/core/handlers/base.py" in get_response
  147.                     response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/views/generic/base.py" in view
  68.             return self.dispatch(request, *args, **kwargs)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/views/generic/base.py" in dispatch
  88.         return handler(request, *args, **kwargs)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/views/generic/edit.py" in post
  256.         return super(BaseCreateView, self).post(request, *args, **kwargs)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/views/generic/edit.py" in post
  222.             return self.form_valid(form)

File "/home/jwe/piesup2/reports/views.py" in form_valid
  182.         return super(ServerCreateView, self).form_valid(form)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/contrib/messages/views.py" in form_valid
  11.         response = super(SuccessMessageMixin, self).form_valid(form)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/views/generic/edit.py" in form_valid
  201.         self.object = form.save()

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/forms/models.py" in save
  452.             self._save_m2m()

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/forms/models.py" in _save_m2m
  434.                 f.save_form_data(self.instance, cleaned_data[f.name])

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/db/models/fields/related.py" in save_form_data
  1618.         setattr(instance, self.attname, data)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/db/models/fields/related_descriptors.py" in __set__
  481.         manager.set(value)

File "/home/jwe/piesup2/venv/lib/python3.4/site-packages/django/db/models/fields/related_descriptors.py" in set
  882.                     (opts.app_label, opts.object_name)

Exception Type: AttributeError at /server-create/
Exception Value: Cannot set values on a ManyToManyField which specifies an intermediary model. Use reports.ServerProduct's Manager instead.

回溯显示,在
form\u valid
方法中调用
super()
时发生错误

File "/home/jwe/piesup2/reports/views.py" in form_valid
   182.         return super(ServerCreateView, self).form_valid(form)
您已经在
form\u valid
方法中保存表单,因此无需调用
super()
。只需重定向到成功url即可

    def form_valid(self, form):
        server = form.save(False)
        server.save()
        for product in form.cleaned_data['products']:
            ServerProduct.objects.create(server=server, product=product)

        return HttpResponseRedirect(self.get_success_url())
请记住添加导入:

from django http import HttpResponseRedirect

请显示
ServerForm
和完整的回溯。请注意,您只需要定义
Server.products
Product.servers
-Django将使用
相关名称创建反向关系。感谢您的建议,我留下了明确的定义,以帮助我理解模型的功能。但是我不认为现在把它们留在那里有什么好处,所以我要把它们拿走:)它一直盯着我的脸!非常感谢你的帮助