Django：在模板中使用Slug的问题

Question

我是Django的初学者。我正在构建一个名为PhoneReview的Django应用。它将存储与最新手机有关的评论。它还将显示电话品牌以及相关的电话型号。

我已经创建了模型，视图和模板文件。现在，我面临一个问题。我不能在网址中使用Slug。现在，它看起来像这样：

这是我的models.py的代码，位于PhoneReview文件夹内：

from django.db import models
from django.template.defaultfilters import slugify

# Create your models here.
class Brand(models.Model):
    brand_name = models.CharField(max_length=100)
    origin = models.CharField(max_length=100)
    manufacturing_since = models.CharField(max_length=100, null=True, blank=True)
    slug = models.SlugField(max_length=150, null=True, blank=True)

    def __str__(self):
        return self.brand_name

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

class PhoneModel(models.Model):
    brand = models.ForeignKey(Brand, on_delete=models.CASCADE)
    model_name = models.CharField(max_length=100)
    launch_date = models.CharField(max_length=100)
    platform = models.CharField(max_length=100)
    slug = models.SlugField(max_length=150, null=True, blank=True)

    def __str__(self):
        return self.model_name

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

class Review(models.Model):
    phone_model = models.ManyToManyField(PhoneModel, related_name='reviews')
    review_article = models.TextField()
    date_published = models.DateField(auto_now=True)
    # slug = models.SlugField(max_length=150, null=True, blank=True)
    link = models.TextField(max_length=150, null=True, blank=True)

    def __str__(self):
        return self.review_article

这是我的urls.py的代码，位于PhoneReview文件夹内：

from . import views
from django.urls import path

urlpatterns = [
    path('index', views.BrandListView.as_view(), name='brandlist'),
    path('phonemodel/<int:pk>/', views.ModelView.as_view(), name='modellist'),
    path('details/<int:pk>/', views.ReviewView.as_view(), name='details'),
]

这是我的views.py的代码，位于PhoneReview文件夹内：

from django.views import generic
from .models import Brand, PhoneModel, Review


class BrandListView(generic.ListView):
    template_name = 'PhoneReview/index.html'
    context_object_name = 'all_brands'

    def get_queryset(self):
        return Brand.objects.all()


class ModelView(generic.DetailView):
    model = PhoneModel
    template_name = 'PhoneReview/phonemodel.html'

class ReviewView(generic.DetailView):
    model = Review
    template_name = 'PhoneReview/details.html'

这是我的apps.py的代码，位于PhoneReview文件夹内：

from django.apps import AppConfig


class PhonereviewConfig(AppConfig):
    name = 'PhoneReview'

这是我的index.html的代码，位于templates文件夹内：

{% extends 'PhoneReview/base.html' %}

{% load static %}

{% block title%}
Brand List
{% endblock %}

{% block content %}
<!--Page content-->
<h1>This is Brand List Page</h1>
<h2>Here is the list of the brands</h2>
    <ul>
        {% for brand in all_brands %}
<!--            <li>{{ brand.brand_name }}</li>-->
            <li><a href = "{% url 'modellist' brand.id %}">{{ brand.brand_name }}</a></li>

        {% endfor %}
    </ul>
<img src="{% static "images/brandlist.jpg" %}" alt="Super Mario Odyssey" /> <!-- New line -->
{% endblock %}

这是我的phonemodel.html的代码，位于templates文件夹内：

{% extends 'PhoneReview/base.html' %}

{% load static %}

{% block title%}
Phone Model Page
{% endblock %}

{% block content %}
<!--Page content-->
<h1>This is Phone Model Page</h1>
<h2>Here is the phone model</h2>
    <ul>
        <li><a href = "{% url 'details' phonemodel.id %}">{{ phonemodel.model_name }}</a></li>
    </ul>
<img src="{% static "images/brandlist.jpg" %}" alt="Super Mario Odyssey" /> <!-- New line -->
{% endblock %}

在index.html中，我尝试将<li><a href = "{% url 'details' phonemodel.id %}">{{ phonemodel.model_name }}</a></li>替换为<li><a href = "{% url 'details' phonemodel.slug %}">{{ phonemodel.model_name }}</a></li>。但是我得到了错误。

如何解决此问题？

Answer 1

如果您要使用Slug，则需要通过URL传递slug而不是pk

path('phonemodel/<slug>/', views.ModelView.as_view(), name='modellist'),
path('details/<slug>/', views.ReviewView.as_view(), name='details'),

现在在模板中，而不是id中

<li><a href = "{% url 'details' phonemodel.slug %}">{{ phonemodel.model_name }}</a></li>