Phi*_*ner 1 django admin class-based-views
我使用基于类的视图创建了一些表单,现在我想将它们添加到Django Admin站点.我刚刚找到此链接,其中描述了向管理站点添加普通视图.
假设您有以下基于类的视图:
# File: views.py
class MyAwesomeBookView(TemplateView):
pass
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要使用此视图,您必须进行此操作 callable
# File: views.py
awesome_book_view = MyAwesomeBookView.as_view()
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一旦你的观点出现callable,你几乎可以像对待基于函数的视图一样对待它.
要将其链接到管理网址,您可以执行以下操作:
# File: admin.py
@admin.register(Book)
class BookModelAdmin(admin.ModelAdmin):
def get_urls(self):
urls = super().get_urls()
my_urls = [
url(r'^awesome-books/$',
self.admin_site.admin_view(awesome_book_view)),
]
return my_urls + urls
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为了与 Django 管理模板完全集成,您可以将模型管理作为额外参数传递给基于类的视图,然后使用它向上下文添加一些糖:
文件管理.py:
from django.contrib import admin
from .models import MyModel
from .views import ProcessObjectView
@admin.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
def get_urls(self):
info = self.model._meta.app_label, self.model._meta.model_name
urls = super().get_urls()
my_urls = patterns('',
url(r'^(?P<object_id>.*)/process/$',
self.admin_site.admin_view(ProcessObjectView.as_view()),
{'model_admin': self, },
name="%s_%s_process" % info),
)
return my_urls + urls
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文件views.py
from django.contrib.auth import get_permission_codename
class ProcessObjectView(UpdateView):
model = MyModel
pk_url_kwarg = "object_id"
fields = [... ]
template_name = 'admin/backend/mymodel/process_object.html'
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
# see http://www.slideshare.net/lincolnloop/customizing-the-django-admin
model_admin = self.kwargs['model_admin']
opts = model_admin.model._meta
admin_site = model_admin.admin_site
has_perm = self.request.user.has_perm(opts.app_label + '.' + get_permission_codename('change', opts))
context.update({
'admin_site': admin_site.name,
'title': 'Process: ' + str(self.get_object()),
'opts': opts,
'app_label': opts.app_label,
'has_chage_permission': has_perm,
})
return context
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文件process_object.html:
{% extends "admin/change_form.html" %}
{% load i18n utils_tags %}
{% block content %}
...
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