Question

我正在尝试使用Django构建一个简单的列表应用程序。我使用的是语言数据类型和具有语言外键的技术数据类型。我希望显示与该语言下的语言相关的所有技术。这是我目前的设置：

models.py

from __future__ import unicode_literals
from django.db import models
class Language(models.Model):
    name = models.CharField(max_length = 32)
    def __str__(self):
        return self.name

class Technology(models.Model):
    class Meta:
        verbose_name_plural = 'technologies'
    name = models.CharField(max_length = 32)
    description = models.TextField()
    language = models.ForeignKey(Language)
    def __str__(self):
        return self.name

views.py：

def skills(request):
    return render(request, 'personal/skills.html')

urls.py：

from django.conf.urls import url, include
from . import views
from django.views.generic import ListView
from personal.models import Language, Technology
urlpatterns = [
    url(r'^$', views.index, name='index'), # Defining url patterns. As things stand, we have a start and an end
    url(r'^contact/', views.contact, name='contact'),
    url(r'^skills/$', ListView.as_view(queryset=Technology.objects.all(), template_name='personal/skills.html'))
]

skills.html：

{%extends 'personal/header.html'%}
{%block content%}
  {%for technology in object_list%}
    <h3>{{technology.name}}</h3>
    <h5>{{technology.language}}</h5>
  {%endfor%}
{%endblock%}

我认为我需要将技能作为URLS文件中的第二个查询集传递给模板，但我不太确定。我应该如何处理这些相关数据集，让它们以我想要的方式返回和显示？

Answer 1

您可以在QuerySet中返回语言，然后从那里访问技术：

q = Language.objects.all().prefetch_related('technology_set')

...
url(r'^skills/$', ListView.as_view(queryset=q, template_name='personal/skills.html'))

在你的模板中：

{% block content %}
  {%for lang in object_list %}
    <h3>{{ lang.name }}</h3>
    {% for tech in lang.technology_set.all %} 
        <h5>{{ tech.name }}</h5>
    {% endfor %}
  {% endfor %}
{% endblock %}

如何列出与Django中的项目关联的所有外键

1 个答案: