Django 动态页面。为什么我的代码不起作用？

Question

我正在创建一个用于销售商品的网站，并且我根据数据库中的数据为每种产品创建一张卡片。当新卡片出现时，基于相同数据库数据的动态页面应该自动连接到它。页面已创建，但模型中的数据尚未加载。我看了 Django 指南，它使用 DetailView 展示了这个方法，我从那里获取了代码。请帮助我，我不明白问题是什么。

My code:

views.py:

class car(DetailView):
    model = inventory
    template_name = 'main/cars.html'
    context_object_name = 'inventory'

urls.py:

from . import views

urlpatterns = [
    path('inventory/<int:pk>', views.car.as_view(), name='cars'),
]

cars.html:

{% extends 'main/layout.html' %}

{% block main %}
{% for el in invent %}
        <div class="main-card ">
            <img  src = '{{ el.img_1 }}' style="">
            <h3 style="">{{ el.name }}</h3>
            <h4 style="">{{ el.rent }}</h4>
            <button><a href="{% url 'cars' el.id %}">Details</a></button>
        </div>
{% endfor %}
{% endblock %}

models.py:

class inventory(models.Model):
    name = models.CharField('Name', max_length=100)
    type = models.CharField('Type of car', max_length=6)

    img_1 = models.ImageField(upload_to='./sql_imgs')
    img_2 = models.ImageField(upload_to='./sql_imgs')
    img_3 = models.ImageField(upload_to='./sql_imgs')
    img_4 = models.ImageField(upload_to='./sql_imgs')
    img_5 = models.ImageField(upload_to='./sql_imgs')
    img_6 = models.ImageField(upload_to='./sql_imgs')
    img_7 = models.ImageField(upload_to='./sql_imgs')
    img_8 = models.ImageField(upload_to='./sql_imgs')

    MSRP = models.CharField('msrp', max_length=40)
    Purchase = models.CharField('purchase', max_length=40)
    rent = models.CharField('rent', max_length=40)

    specs = models.TextField('specs')

    text = models.TextField('About car')


    def str(self):
        return self.name

    class Meta:
        verbose_name = 'Inventory'
        verbose_name_plural = 'Inventory'

我读了很多文章，但没有发现任何有价值的东西。希望我能在这里找到答案。

Answer 1

请像这样更新您的代码：

models.py

from django.db import models

class Inventory(models.Model):
    name = models.CharField('Name', max_length=100)
    type = models.CharField('Type of car', max_length=6)
    img_1 = models.ImageField(upload_to='sql_imgs/')
    img_2 = models.ImageField(upload_to='sql_imgs/')
    img_3 = models.ImageField(upload_to='sql_imgs/')
    img_4 = models.ImageField(upload_to='sql_imgs/')
    img_5 = models.ImageField(upload_to='sql_imgs/')
    img_6 = models.ImageField(upload_to='sql_imgs/')
    img_7 = models.ImageField(upload_to='sql_imgs/')
    img_8 = models.ImageField(upload_to='sql_imgs/')
    MSRP = models.CharField('msrp', max_length=40)
    Purchase = models.CharField('purchase', max_length=40)
    rent = models.CharField('rent', max_length=40)
    specs = models.TextField('specs')
    text = models.TextField('About car')

    def __str__(self):
        return self.name

    class Meta:
        verbose_name = 'Inventory'
        verbose_name_plural = 'Inventories'

views.py

from django.views.generic.detail import DetailView
from .models import Inventory

class CarDetailView(DetailView):
    model = Inventory
    template_name = 'main/cars.html'
    context_object_name = 'inventory'

urls.py

from django.urls import path
from .views import CarDetailView

urlpatterns = [
    path('inventory/<int:pk>/', CarDetailView.as_view(), name='cars'),  # Added trailing slash for best practice
]

cars.html

{% extends 'main/layout.html' %}

{% block main %}
<div class="main-card">
    <img src="{{ inventory.img_1.url }}" alt="{{ inventory.name }}">
    <h3>{{ inventory.name }}</h3>
    <h4>{{ inventory.rent }}</h4>
    <p>{{ inventory.text }}</p>
</div>
{% endblock %}

请像这样更新您的代码，让我们看看接下来会发生什么来根据需要修复它。