【发布时间】:2018-03-30 15:37:28
【问题描述】:
我正在尝试设置我的网站,以便我的工作详细信息的 url 将使用 slug 字段而不是 pk。它告诉我它无法使用给定的 slug(它是一个 int,147)找到我的工作。
更新:
查看https://ccbv.co.uk/projects/Django/1.11/django.views.generic.detail/DetailView/ 的DetailView 描述后,我意识到DetailView 有一个slug_field 属性。我的新视图如下所示:
class JobDetailView(CacheMixin, DetailView):
model = Job
slug_field = 'slug'
问题:
网址:
urlpatterns = [
url(r'^careers$', views.job_list, name='job-list'),
url(r'^careers/(?P<slug>[0-9]+)/$', views.JobDetailView.as_view(), name='job-detail'),
]
查看:
class JobDetailView(CacheMixin, DetailView):
model = Job
pk_url_kwarg = 'slug'
def get_object(self, *args, **kwargs):
# Call the superclass
object = super(JobDetailView, self).get_object()
# Return the object
return object
def get(self, request, *args, **kwargs):
object = super(JobDetailView, self).get(request, *args, **kwargs)
return object
型号:
class Job(UpdateAble, PublishAble, models.Model):
slug = models.CharField(unique=True, max_length=25)
facility = models.ForeignKey('Facility')
recruiter = models.ForeignKey('Recruiter')
title = models.TextField()
practice_description = models.TextField(blank=True, default="")
public_description = models.TextField(blank=True, default="")
objects = JobManager()
def get_next(self, **kwargs):
jobs = Job.objects.published()
next = next_in_order(self, qs=jobs)
if not next:
next = jobs[0]
return next
def get_prev(self, **kwargs):
jobs = Job.objects.published()
prev = prev_in_order(self, qs=jobs)
if not prev:
prev = jobs[len(jobs)-1]
return prev
def __str__(self):
return f'{self.facility}; {self.title}'
经理:
class JobManager(models.Manager):
def published(self):
return super(JobManager, self).get_queryset().filter(is_published=True).order_by('facility__name', 'title')
【问题讨论】:
-
为什么你的 slug 字段是 int ?这违背了拥有蛞蝓的全部目的。另请注意,您的
get_object和get方法都是没有意义的,您应该删除它们。 -
这是一个 int,因为我使用来自其他系统的 ID 作为我的 slug。我有充分的理由这样做。 2)我曾一度使用覆盖这些方法来实现某些目标,但没有注意之后不需要覆盖。我会删除它们。
标签: django django-views django-urls django-class-based-views