具有多个网址的站点地图和对象

Sitemap and object with multiple urls

站点地图在Django中的正常使用方式是:

from django.contrib.sitemaps import Sitemap
from schools.models import School


class SchoolSitemap(Sitemap):
    changefreq = "weekly"
    priority = 0.6

    def items(self):
        return School.objects.filter(status = 2)

然后在School的模型中我们定义:

  def get_absolute_url(self):
      return reverse('schools:school_about', kwargs={'school_id': self.pk})

在这样的实施中,我在 sitemap.xml

的一所学校有一个关于 link

问题是我的学校有多个页面:About、Teachers、Pupils 和其他,我希望呈现的所有页面都是 sitemap.xml

最好的方法是什么?

你可以处理 items may return anything that can be passed to the other methods of a Sitemap:

import itertools

class SchoolSitemap(Sitemap):
    # List method names from your objects that return the absolute URLs here
    FIELDS = ("get_absolute_url", "get_about_url", "get_teachers_url")

    changefreq = "weekly"
    priority = 0.6

    def items(self):
        # This will return you all possible ("method_name", object) tuples instead of the
        # objects from the query set. The documentation says that this should be a list 
        # rather than an iterator, hence the list() wrapper.
        return list(itertools.product(SchoolSitemap.FIELDS,
                                      School.objects.filter(status = 2)))

    def location(self, item):
        # Call method_name on the object and return its output
        return getattr(item[1], item[0])()

如果字段的数量和名称没有预先确定,我会采用完全动态的方法:允许模型有一个 get_sitemap_urls 方法,returns 一个绝对 URL 列表,并使用执行此方法的 Sitemap。也就是说,在最简单的情况下,您不需要访问 priority/changefreq/lastmod 方法中的对象:

class SchoolSitemap(Sitemap):
    changefreq = "weekly"
    priority = 0.6

    def items(self):
        return list(
             itertools.chain.from_iterable(( object.get_sitemap_urls()
                                             for object in 
                                             School.objects.filter(status = 2)))
        )

    def location(self, item):
        return item