Django DetailView 按外键过滤答案

【问题标题】：Django DetailView Filter by Foreign KeyDjango DetailView 按外键过滤
【发布时间】：2012-12-15 01:47:46
【问题描述】：

我有点困惑，想利用 DetailView 功能，使用外键作为我的过滤器来显示数据。基本上我的模型是这样的：

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 %}

基本上我想在我的 category_detail.html 中显示以下信息：

类别名称

书名

出版商名称

发布日期

任何帮助将不胜感激。

谢谢你-Keoko

【问题讨论】：

标签： django foreign-keys detailview

【解决方案1】：

感谢您的友好回复。我创建了一个包含以下信息的 views.py 文件：

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()),  
)

最后修改 category_detail.html 如下：

{% 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 %}

【讨论】：

【解决方案2】：

在模板中你有类别对象。您可以遍历所有书籍；

{% 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 }}

【讨论】：