Question

我正在尝试允许用户使用UpdateView编辑输入到项目中的信息。我一直收到错误说＆＃34;通用详细信息视图必须使用对象pk或slug调用UpdateProject。＆＃34;我已尝试指向UpdateView类中的slug，但无法使其工作。我是Django的新手，所以任何帮助都会非常感激！

视图

class CreateProject(CreateView):
    model = UserProject
    form_class = UserProjectForm
    template_name = 'howdidu/create_project.html'

    def form_valid(self, form):
        userproject = form.save(commit=False)
        form.instance.user = self.request.user
        userproject.save()
        self.object = userproject
        return super(CreateProject, self).form_valid(form)

    def get_success_url(self):
        project_username = self.request.user.username
        project_slug = self.object.slug
        return reverse('user_project', kwargs={'username':project_username, 'slug': project_slug})

class UpdateProject(UpdateView):
    model = UserProject
    form_class = UserProjectForm
    template_name = 'howdidu/update_project.html'
    slug_field = 'slug'

    def form_valid(self, form):
        userproject = form.save(commit=False)
        form.instance.user = self.request.user
        userproject.save()
        self.object = userproject
        return super(UpdateProject, self).form_valid(form)

    def get_success_url(self):
        project_username = self.request.user.username
        project_slug = self.object.slug
        return reverse('user_project', kwargs={'username':project_username, 'slug': project_slug})

网址

urlpatterns = patterns('',
        url(r'^$', views.index, name='index'),
        url(r'^register_profile/$', views.register_profile, name='register_profile'),
        url(r'^update_profile/$', views.update_profile, name='update_profile'),
        url(r'^create_project/$', login_required(views.CreateProject.as_view()), name='create_project'),
        url(r'^update_project/$', login_required(views.UpdateProject.as_view()), name='update_project'),
        url(r'^(?P<username>\w+)/$', views.profile_page, name='user_profile'),
        url(r'^(?P<username>\w+)/(?P<slug>[-\w]+)/$', views.project_page, name='user_project'),
        )

模型

class UserProject(models.Model):
    user = models.ForeignKey(User)
    title = models.CharField(max_length=100)
    project_overview = models.CharField(max_length=1000)
    project_picture = models.ImageField(upload_to='project_images', blank=True)
    date_created = models.DateTimeField(auto_now_add=True)
    project_views = models.IntegerField(default=0)
    project_likes = models.IntegerField(default=0)
    project_followers = models.IntegerField(default=0)
    slug = models.SlugField(max_length=100, unique=True) #should this be unique or not?

    def save(self, *args, **kwargs):
        self.slug = slugify(self.title)
        super(UserProject, self).save(*args, **kwargs)

    def __unicode__(self):
        return self.title

Answer 1

Django的UpdateView默认需要来自相应网址的pk或slug参数。在您的情况下，您需要将URL更改为UpdateProject视图

url(r'^update_project/(?P<slug>[-\w]+)/$',
    login_required(views.UpdateProject.as_view()),
    name='update_project'),

slug是你的项目的slu ..

Answer 2

在您的应用中创建utils.py文件。使用django utils中的slugify。

import random
import string


from django.utils.text import slugify



def random_string_generator(size=10, chars=string.ascii_lowercase + string.digits):
    return "".join(random.choice(chars) for _ in range(size))


def unique_slug_generator(instance, new_slug=None):
    if new_slug is None:
        slug = new_slug
    else:
        slug=slugify(instance.model field of your choice)
    Klass = instance.__class__
    qs_exists = Klass.objects.filter(slug=slug).exists()
    if qs_exists:
        new_slug = "{slug}-{randstr}".format(
            slug=slug,
            randstr=random_string_generator(size=4)
            )
        return unique_slug_generator(instance, new_slug=new_slug)
    return slug

向模型添加块

slug = models.SlugField(null=True, blank=True)

将其添加到您的urls.py文件中

url(r'^update/(?P<slug>[\w-]+)/$', profileUpdate.as_view(), name='update'),

Answer 3

class UserProject(UpdateView):
    model = BlogSlider
    fields = ['title', 'sub_title', 'description', 'image']
    template_name = 'update.html'

使用表格并使用{{form.as_p}}。

无法将slug传递给UpdateView以便获取正确的数据

3 个答案: