基于url参数的Django CreateView自定义表单默认字段

时间:2013-12-06 09:59:52

标签: python django django-forms create-view

file:Capacity / models.py

class Env(models.Model):
    name = models.CharField(max_length=50)
    def get_absolute_url(self):
            return reverse('index')

class Envhosts(models.Model):
    env =  models.ForeignKey(Env)
    hostname = models.CharField(max_length=50)
    count = models.IntegerField()

    class Meta:
        unique_together = ("env","hostname")

    def get_absolute_url(self):
        return reverse('index')

file:Capacity / views.py

 class EnvhostsCreate(CreateView):
    model = Capacity.models.Envhosts
    fields=['env','hostname','count']
    template_name_suffix = '_create_form'

文件Capacity / urls.py:

urlpatterns = patterns(........ url(r'^createhosts/(?P<envid>\d+)/$',EnvhostsCreate.as_view(),name='envhosts_create'))

现在, 当我打开这个表格时: /Capacity/createhosts/3/(其中3是我的环境) 它根据Env的对象数显示env对象的选项作为下拉列表。但是我希望它能够根据env id(在这种情况下为'3')单独使用env

我知道我必须在类EnvhostsCreate(CreateView)中覆盖一些方法。但我无法确定哪种方法以及如何根据/createhosts/

之后的部分采用env

1 个答案:

答案 0 :(得分:4)

您可以使用the documentation中描述的模式添加request.user - 它的原理相同。从字段列表中删除env,然后定义form_valid()

class EnvhostsCreate(CreateView):
    model = Capacity.models.Envhosts
    fields = ['hostname', 'count']
    template_name_suffix = '_create_form'

    def form_valid(self, form):
        form.instance.env = Envhosts.objects.get(pk=self.kwargs['envid'])
        return super(EnvhostsCreate, self).form_valid(form)