2010-08-25 77 views
2

我试图在我的Django模板中输出以下数据。Django:通过计数排序的聚集输出聚合

国家将按照故事排列顺序排列。 城市将被勒令故事#下降(该国下)

Country A(# of stories) 
    City A (# of stories) 
    City B (# of stories) 

Country B(# of stories) 
    City A (# of stories) 
    City B (# of stories) 

我的模型如下:

# Create your models here. 
class Country(models.Model): 
    name = models.CharField(max_length=50) 

class City(models.Model): 
    name = models.CharField(max_length=50) 
    country = models.ForeignKey(Country) 

class Story(models.Model): 
    city = models.ForeignKey(City) 
    country = models.ForeignKey(Country) 
    name = models.CharField(max_length=255) 

什么是最简单的方法是什么?

+0

我意识到是这样做的方法很多,但不知道是否有利用Django的ORM的方式,像这样的答案在这里,http://stackoverflow.com/questions/ 1010848 /遍历多维字典在Django中/ 1011145#1011145 – Maverick 2010-08-25 09:33:50

回答

0

此解决方案适用于我。你需要调整它才能将它传递给模板。

from django.db.models import Count 
all_countries = Country.objects.annotate(Count('story')).order_by('-story__count') 

for country in all_countries: 
    print "Country %s (%s)" % (country.name, country.story__count) 
    all_cities = City.objects.filter(country = country).annotate(Count('story')).order_by('-story__count') 
    for city in all_cities: 
     print "\tCity %s (%s)" % (city.name, city.story__count) 

更新

这里是将这些信息发送到模板的一种方式。这一个涉及使用自定义过滤器。

@register.filter 
def get_cities_and_counts(country): 
    all_cities = City.objects.filter(country = country).annotate(Count('story')).order_by('-story__count') 
    return all_cities 

查看:

def story_counts(request, *args, **kwargs): 
    all_countries = Country.objects.annotate(Count('story')).order_by('-story__count') 
    context = dict(all_countries = all_countries) 
    return render_to_response(..., context) 

而且在你的模板:

{% for country in all_countries %} 
    <h3>{{ country.name }} ({{ country.story__count }})</h3> 
    {% for city in country|get_cities_and_counts %} 
     <p>{{ city.name }} ({{ city.story__count }})</p> 
    {% endfor %} 
{% endfor %} 

更新2

变异与模型的自定义方法。

class Country(models.Model): 
    name = models.CharField(max_length=50) 

    def _get_cities_and_story_counts(self): 
     retrun City.objects.filter(country = self).annotate(Count('story')).order_by('-story__count') 
    city_story_counts = property(_get_cities_and_story_counts) 

这可以让您避免定义过滤器。模板代码更改为:

{% for country in all_countries %} 
    <h3>{{ country.name }} ({{ country.story__count }})</h3> 
    {% for city in country.city_story_counts %} 
     <p>{{ city.name }} ({{ city.story__count }})</p> 
    {% endfor %} 
{% endfor %} 
+0

谢谢,但有没有一种自然的方式将此传递到我可以在模板中工作的数据结构?或者我现在需要建立一个自定义字典? – Maverick 2010-08-25 07:24:32

+0

增加了一种方法。这涉及一个过滤器。 – 2010-08-25 07:36:49

+0

我喜欢这个答案,但想知道是否有更“django”这样做的方式。像这样的答案,http://stackoverflow.com/questions/1010848/traversing-multi-dimensional-dictionary-in-django/1011145#1011145 – Maverick 2010-08-25 09:32:34