我的Django应用程序之一"Pages"在models.py中有许多不同的类。每个类都有自己的slug属性/字段。我可以加载第一个"公司"(第一类(的URL，但由于某些原因无法加载其他URL。
我尝试过分别导入视图，并为每个视图创建url路径。这样做时，只会加载最后一个URL(第三个(。
型号.py
class Industry(models.Model):
industry = models.CharField(max_length=140, null=True, blank=True, unique=True)
slug = models.SlugField(max_length=40, null=True, blank=True)
def __str__(self):
return self.industry
def save(self, *args, **kwargs):
if not self.id:
self.slug = slugify(self.industry)
super(Industry, self).save(*args, **kwargs)
def get_absolute_url(self):
return reverse('industry_detail', args=[(self.slug)])
class Subindustry(models.Model):
subindustry = models.CharField(max_length=140, null=True, blank=True, unique=True)
industry = models.ForeignKey(
Industry,
on_delete=models.CASCADE,
related_name='ParentIndustry',
)
slug = models.SlugField(max_length=40, null=True, blank=True)
def __str__(self):
return self.subindustry
def save(self, *args, **kwargs):
if not self.id:
self.slug = slugify(self.subindustry)
super(Subindustry, self).save(*args, **kwargs)
def get_absolute_url(self):
return reverse('subindustry_detail', args=[(self.slug)])
class Company(models.Model):
name = models.CharField(max_length=50, blank=False, unique=True, default=(str(id)))
website = models.URLField(max_length=100)
...
slug = models.SlugField(max_length=40, null=True, blank=True)
def __str__(self):
return self.name
def save(self, *args, **kwargs):
if not self.id:
self.slug = slugify(self.name)
super(Company, self).save(*args, **kwargs)
def get_absolute_url(self):
return reverse('company_detail', args=[(self.slug)])
视图.py
from django.views.generic import DetailView
from django.urls import reverse_lazy
# Create your views here.
from .models import Company, Industry, Subindustry
class CompanyDetailView(DetailView):
model = Company
template_name = 'company_detail.html'
slug_field = 'slug'
class IndustryDetailView(DetailView):
model = Industry
slug_field = 'slug'
template_name = 'industry_detail.html'
class SubindustryDetailView(DetailView):
model = Subindustry
slug_field = 'slug'
template_name = 'subindustry_detail.html'
urls.py(页面应用程序(
from django.urls import path
from .views import CompanyDetailView, IndustryDetailView, SubindustryDetailView
urlpatterns = [
path('<slug:slug>/', CompanyDetailView.as_view(), name='company_detail'),
path('<slug:slug>/', IndustryDetailView.as_view(), name='industry_detail'),
path('<slug:slug>/', SubindustryDetailView.as_view(), name='subindustry_detail'),
]
urls.py(项目级(
from django.contrib import admin
from django.urls import path, include
from django.views.generic.base import TemplateView
urlpatterns = [
path('admin/', admin.site.urls),
path('users/', include('users.urls')),
path('users/', include('django.contrib.auth.urls')),
path(r'aoa/', include('qa.urls')),
path('business/', include('pages.urls')),
path('', TemplateView.as_view(template_name='home.html'), name='home'),
]
预期：输入正确的url时看到页面。例如，如果我尝试加载http://127.0.0.1:8000/business/construction/它将加载(例如，这是Industry值(。目前只有公司重视负载，例如。http://127.0.0.1:8000/business/macrodepot/请注意，我确实有每个类的HTML模板，如果我在views.py文件中只有一个模板，我就可以看到它们。
实际：当我输入行业或子行业的url时，会收到下面的404。就好像只在公司的详细信息中寻找鼻涕虫。
找不到页面(404(请求方法：GET请求URL:http://127.0.0.1:8000/business/construction/引发者：pages.views.CompanyDetailView找不到与查询匹配的公司
这是因为django按照url在url模式中的放置顺序来评估url。它接收到对企业/建筑的请求，并调用CompanyDetailView，返回404。您需要为每个模型提供不同的url模式：
urlpatterns = [
path('company/<slug:slug>/', CompanyDetailView.as_view(), name='company_detail'),
path('industry/<slug:slug>/', IndustryDetailView.as_view(), name='industry_detail'),
path('subindustry/<slug:slug>/', SubindustryDetailView.as_view(), name='subindustry_detail'),
]