Question

我很困惑，并希望使用外键作为我的过滤器来显示数据的DetailView功能。基本上我的模型看起来像这样：

class Category(models.Model):
    name = models.CharField(max_length=30)
    slug = models.SlugField(help_text="A short name for Category")

    def __unicode__(self):
       return self.name

    class Meta:
       ordering = ["-name"]
       verbose_name_plural = "categories"  


class Publisher(models.Model):
    name = models.CharField(max_length=30)
    slug = models.SlugField(help_text="A short name for Publisher")

    class Meta:
       ordering = ["-name"]

    def __unicode__(self):
        return self.name


class Book(models.Model):
    title = models.CharField(max_length=100)
    slug = models.SlugField(help_text="A short name for book")
    pub_date = models.DateField()
    publisher = models.ForeignKey(Publisher)
    category = models.ForeignKey(Category)

    class Meta:
       ordering = ["-title"]

    def __unicode__(self):
       return self.title

我的Urls.py：

url(r'^categories/(?P<slug>[\w-]+)/$', DetailView.as_view(model=Category,template_name="books/category_detail" )),

我的Category_detail.html

{% block content %}
    <h2>{{ category.name}} </h2>
    <ul>       
    <li>Book Title: {{ book.title }}</li>
<li>Book publisher: {{ book.publisher }}</li>
    <li>Book Published Date: {{ book.pub_date }}</li>    
    </ul>
{% endblock %}

基本上我想在category_detail.html中显示以下信息：

类别名称

书名

发布商名称

发布日期

任何帮助都将不胜感激。

谢谢 - Keoko

Answer 1

感谢您的回复。我创建了一个views.py文件，其中包含以下信息：

from django.shortcuts import get_object_or_404
from django.views.generic import ListView
from mysite.books.models import *

class BooksCategoryListView(ListView):

   context_object_name = "book_list"

   "get_queryset = query all the objects in the database"
    def get_queryset(self):
      category_slug = get_object_or_404(Category, slug=self.kwargs['slug'])
      return Book.objects.filter(category=category_slug)

并更新了我的应用程序urls.py:

from django.conf.urls import patterns, url, include
from django.views.generic import ListView, DetailView
from mysite.books.views import BooksCategoryListView
from mysite.books.models import *

urlpatterns = patterns('',
   ...snip....        
   url(r'^categories/(?P<slug>[\w-]+)/$', BooksCategoryListView.as_view()),  
)

最后使用以下内容修改了category_detail.html：

{% block content %}
<h2>Book Details</h2>
<ul>     
   <li>Category: {{ book.category}}</li> 
   <li>Title: {{ book.title }}</li>
   <li>Author: {{ book.authors }}</li>
   <li>Publisher: {{ book.publisher }}</li>
   <li>Published Date: {{ book.pub_date }}</li>            
</ul>
{% endblock %}

Answer 2

在模板中，您有类别对象。你可以遍历书籍;

{% for book in category.book_set.all %}
   Book Title: {{ book.title }}
   Book publisher: {{ book.publisher }}
   Book Published Date: {{ book.pub_date }}
{% endfor %}

或者只是第一本书;

   Book Title: {{ category.book_set.all.0.title }}
   Book publisher: {{ category.book_set.all.0.publisher }}
   Book Published Date: {{ category.book_set.all.0.pub_date }}

Django DetailView按外键过滤

2 个答案: