python - Django Forms，为 request.user 设置一个初始值

Question

有什么方法可以使以下内容成为可能，还是应该在其他地方完成？

class JobRecordForm(forms.ModelForm):
    supervisor = forms.ModelChoiceField(
        queryset    = User.objects.filter(groups__name='Supervisors'), 
        widget      = forms.RadioSelect,
        initial     = request.user # is there some way to make this possible?
    )    
    class Meta:
        model = JobRecord

Answer 1

如果您在 view.py 中执行此操作：

form = JobRecordForm( initial={'supervisor':request.user} )

然后你不会触发验证。

请参阅http://docs.djangoproject.com/en/dev/ref/forms/api/#dynamic-initial-values

Answer 2

您可能希望在视图函数中处理此问题。由于您的视图函数必须创建初始表单，并且您的视图函数知道用户。

form = JobRecordForm( {'supervisor':request.user} )

这将触发对此输入的验证，顺便说一句，因此您不能以这种方式提供提示值。

Answer 3

另一个使用中间件的解决方案并节省重写：使用中间件解决方案您可以在任何地方调用“请求”。

---

"""中间件"""

    # coding: utf-8 
from django.utils.thread_support import currentThread 
_requests = {}

def get_request():
    return _requests[currentThread()]

class GlobalRequestMiddleware(object):
    def process_request(self, request):  
        _requests[currentThread()] = request

---

"""保存重写"""

class Production(models.Model):
    creator = models.ForeignKey(User, related_name = "%(class)s_creator")
    creation_date = models.DateTimeField(auto_now_add = True)
    modification_date = models.DateTimeField(auto_now = True)

    def save(self, force_insert = False, force_update = False):

        self.creator = get_request().user
        super(Production, self).save(force_insert = force_insert, force_update = force_update)
        return

Answer 4

要获得完整的答案，这里是 CBV 解决方案：

class MyFormView(TemplateView, FormMixin):
    def get_initial(self):
        self.initial.update({'your_field': self.request.user})
        return super(MyFormView, self).get_initial()