使用 slugs 和基于 class 的视图时找不到视图

Failing to find views when using slugs and class based views

我的一个 Django 应用程序 "Pages" 在 models.py 中有许多不同的 classes。他们每个人都有自己的鼻涕虫attribute/field。我能够加载第一个 URL - "company"(第一个 class)但由于某种原因无法加载其他 URLs.

我试过单独导入视图并为每个视图创建 urls 路径。执行此操作时,只会加载最后的 URLs(第三个)。

models.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)])

views.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)

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(项目级别)

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:向右输入时看到页面url。例如。如果我尝试加载 http://127.0.0.1:8000/business/construction/ it will load (this is an Industry value for example). Currently only company values load, e.g. http://127.0.0.1:8000/business/macrodepot/ 请注意,我确实有每个 class 的 HTML 模板,如果我在 views.py 文件中只有其中一个模板,我就能看到它们。

实际:当我输入某个行业或子行业的 url 时收到以下 404。好像只在 companydetailview 中寻找 slug。

找不到页面 (404) 请求方式:GET 请求 URL:http://127.0.0.1:8000/business/construction/ 提出者:pages.views.CompanyDetailView 找不到符合查询的公司

这是因为 django 按照 url 放在 url 模式中的顺序计算 url。它收到对 businesses/construction 的请求并调用 CompanyDetailView,其中 returns 一个 404。您需要为每个模型设置不同的 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'),
]