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python – Django CreateView基于url参数自定义表单默认字段

作者:互联网

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'

文件容量/ urls.py:

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

所以现在,
当我打开这个表格时:
/ capacity / createhosts / 3 /(其中3是我的环境ID)
它根据Env的对象数显示env对象的选项作为下拉列表.但我希望它基于env id(在这种情况下为’3′)自己采用env

我知道我必须在类EnvhostsCreate(CreateView)中覆盖一些方法.但我无法弄清楚哪种方法以及如何在/ createhosts /之后基于部分获取env

解决方法:

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)

标签:python,django,django-forms,create-view
来源: https://codeday.me/bug/20190624/1281114.html