python - 如何从数据库中获取一个国家的城市列表？

Question

我有 2 个模型，城市和国家。我想根据所选国家/地区在页面上显示该国家/地区的城市。我尝试查询将国家/地区模型传递给城市模型以查找与该国家/地区相关的所有城市，但目前，页面显示所有城市，无论该国家/地区是什么。

当用户从页面中选择国家/地区时，如何显示该国家/地区的城市？

非常感谢任何帮助或建议。

models.py

class City(models.Model):
    country = models.ForeignKey('Country', on_delete=models.CASCADE, related_name='country')
    name = models.CharField(max_length=90, verbose_name='City name')
    slug = models.SlugField(null=True, unique=True)

    def __str__(self):
        return self.name

    def get_absolute_url(self):
        return reverse('city:cities', kwargs={'slug': self.slug})

     def save(self, *args, **kwargs): 
         if not self.slug:
             self.slug = slugify(self.name)
         return super().save(*args, **kwargs)

class Country(models.Model):
    country_name = models.CharField(max_length=50, verbose_name='Country name',unique=True)
    slug = models.SlugField(null=True, unique=True)

    def __str__(self):
        return self.country_name

    def get_absolute_url(self):
       return reverse('city:cities', kwargs={'slug': self.slug})

    def save(self, *args, **kwargs): 
         if not self.slug:
             self.slug = slugify(self.country_name)
         return super().save(*args, **kwargs)

views.py

class CountriesListView(ListView):
    template_name = "countries.html"
    model = Country

    context_object_name = "countries"

class CitiesListView(ListView):
    template_name = "cities_list.html"
    model = City

    context_object_name = "cities"

    def get_context_data(self, *args, **kwargs):
        context = super(CitiesListView, self).get_context_data(*args,**kwargs)
        countries = Country.objects.all()
        cities = City.objects.filter(country__in=countries)
        context['cities'] = cities
        return context

模板

# countries.html

<h1>Countries</h1>

{% for country in countries %}
  <h3><a href="{% url 'city:cities' country.slug %}">{{ country.country_name }}</h3>
{% endfor %}


# cities_list.html

<h1>Cities</h1>

{% for city in cities %}
  <div>
        <h3><a href="{% url 'city:details' city.slug %}">{{ city }}</a></h3>
  </div>  
{% endfor %}

Answer 1

related_name=… parameter [Django-doc]是反向关系的名称，因此可以访问特定Country的City对象。因此将其命名为 Country 并不是一个好主意。您可以将其重命名为cities，例如:

class City(models.Model):
    country = models.ForeignKey(
        'Country',
        on_delete=models.CASCADE,
        <b>related_name='cities'</b>
    )
    # …

您可以使用 DetailView 呈现单个 Country，因此:

from django.views.generic.detail import <b>DetailView</b>

class CountryDetailView(<b>DetailView</b>):
    template_name = 'country_detail.html'
    model = Country
    context_object_name = 'country'

在 URL 中，您可以使用 slug 字段以名称 cities 注册此 View :

# app/urls.py

from django.urls import path

from . import views

app_name='city'

urlpatterns = [
    # …,
    path(
        <b>'country/<slug:slug>'</b>,
        views.CountryDetailView.as_view(),
        <b>name='cities'</b>
    ),
]

在模板中，您可以反向访问关系:

<!-- country_detail.html -->

{{ country }}

{% for <b>city in country.cities.all</b> %}
    {{ city }}
{% endfor %}