基于类的视图查询:获取另一个模型引用的对象 [英] Class based views query: get objects referenced by another model
问题描述
此问题扩展了
Django将多个模型传递给一个模板
我想生成1个模板,即我的个人资料索引页,因此它将列出url <中指定的用户创建的所有对象/ p>
我想查询一个模板的多个模型,获取有问题的概要文件,并显示由附加概要文件创建的所有oferto对象
views.py
class ProfileIndex(ListView):
context_object_name ='account_list'
template_name ='profiles / account_index.html'
queryset = Profile.objects.all()
def get_context_data(self,** kwargs):
context = super (ProfileIndex,自身).get_context_data(** kwargs)
context ['profile'] = Profile.objects.all()
context ['oferto'] = Oferto.objects.get(kwargs.profile .slug)
返回上下文
urls.py我想我需要从url中获取配置文件信息,以便通过网址中的该用户
url(
regex = r ^ index /(?P< slug> [- _\w] +),
view = ProfileIndex.as_view(),
name = profile_index,
),
这是最难找出的,如何使我的模板从相关上下文中正常工作
{%块内容%}
< h1> {{个人资料}}< h1>
< h2> Ofertoj或/aŭOffer< / h2>
{%for aerto in profile%}
< a href = {%url oferto_detail oferto.slug%}> {{oferto.name}}< / a>< br />
{%endfor%}
{%endblock%}
frontend / models.py
Class Profile(models.Model):
....
Class Oferto(models.Model):
profile = model。 ForeignKey(配置文件):
...
frontend / views.py
Class ViewProfileList(ListView):
模型=配置文件
template_name ='frontend / view_profile_list.html'
frontend / templates / frontend / view_profile_list.html
{%为object_list.profile中的配置文件。所有%}
{%为profile.oferto_set.all%中的销售者}
{{ofer.you_field}}
{ %endfor%}
{%endfor%}
this question extends Django Pass Multiple Models to one Template
In that I want to produce 1 template, that is my profile index page, so it will list all the objects created by the user specified in the url
I want to query multiple models for one template, get the Profile in question and show all the oferto objects that was created by the attached profile
views.py
class ProfileIndex(ListView):
context_object_name = 'account_list'
template_name = 'profiles/account_index.html'
queryset = Profile.objects.all()
def get_context_data(self, **kwargs):
context = super(ProfileIndex, self).get_context_data(**kwargs)
context['profile'] = Profile.objects.all()
context['oferto'] = Oferto.objects.get(kwargs.profile.slug)
return context
urls.py I am thinking I need to get the profile slug from the url in order to search for all oferto by that user in the url
url(
regex=r"^index/(?P<slug>[-_\w]+)",
view=ProfileIndex.as_view(),
name="profile_index",
),
And this has been the hardest to figure out, how to make my template work correctly from the context in question
{% block content %}
<h1>{{ profile }}<h1>
<h2> Ofertoj or/aŭ Offers</h2>
{% for oferto in profile %}
<a href="{% url "oferto_detail" oferto.slug %}">{{ oferto.name }}</a><br/>
{% endfor %}
{% endblock %}
frontend/models.py
Class Profile(models.Model):
....
Class Oferto(models.Model):
profile = model.ForeignKey(Profile):
...
frontend/views.py
Class ViewProfileList(ListView):
model = Profile
template_name = 'frontend/view_profile_list.html'
frontend/templates/frontend/view_profile_list.html
{% for profile in object_list.all %}
{% for ofer in profile.oferto_set.all %}
{{ ofer.you_field }}
{% endfor %}
{% endfor %}
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