如何在 CreateView 中为自定义用户模型设置外键？

Question

我有这个自定义用户模型“es_user”

models.py

class es_user(models.Model):

user = models.OneToOneField(User, on_delete=models.CASCADE)

class es_event(models.Model):

    ev_name = models.CharField(max_length=100)

    ev_venue = models.CharField(max_length=100)

    ev_admin =  models.ForeignKey('es_user',related_name='events',on_delete=None)

最终，我将向 es_user 添加更多字段，这就是我使用自定义用户模型的原因。所以我不能满足于 django 的内置用户模型。

views.py

class CreateEventView(LoginRequiredMixin,CreateView):

    model = es_event

    fields = ['ev_name','ev_venue','ev_date','ev_description']

    def form_valid(self, form):

        form.instance.ev_admin = self.request.user

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

当我提交表单时出现此错误

Cannot assign "<SimpleLazyObject: <User: randy>>": "es_event.ev_admin" must be a "es_user" instance.

我已经检查了 Django 文档和其他堆栈溢出帖子，但在所有这些帖子中，外键都引用了 Django 的内置用户模型

Answer 1

所以只分配 es_user，而不是 auth 用户。form.instance.ev_admin&nbsp;=&nbsp;self.request.user.es_user

Answer 2

当我对 views.py 和 models.py 进行一些更改时，一切正常views.pyclass CreateEventView(LoginRequiredMixin,CreateView):model = es_eventfields = ['ev_name','ev_venue','ev_date','ev_description']def form_valid(self, form):&nbsp; &nbsp; form.instance.ev_admin = self.request.user.es_usermodels.pyclass es_user(models.Model):&nbsp; &nbsp; &nbsp; user = models.OneToOneField(User,related_name='es_user', on_delete=models.CASCADE)

Answer 3

你可以es_user取而代之：form.instance.ev_admin&nbsp;=&nbsp;es_user.objects.get(user=self.request.user)