金字塔：如何自动生成站点地图？

Question

Is there a way to generate sitemaps on the fly and regularly submit them to Google using Pyramid?

I've seen 2 code snippets ( here and here ) for doing this in Flask, but they don't seem applicable to Pyramid.

Specifically, when I include config.add_route('sitemap', '/sitemap.xml') in __init__.py and then the following view:

@view_config(route_name='sitemap', renderer='static/sitemap.xml')
def sitemap(request):
    ingredients = [ ingredient.name for ingredient in Cosmeceutical.get_all() ]
    products = [ product.name for product in Product.get_all() ]
    return dict(ingredients=ingredients, products=products)

I get an error:

File "/home/home/SkinResearch/env/local/lib/python2.7/site-packages/pyramid-1.4.5-    py2.7.egg/pyramid/registry.py", line 148, in _get_intrs_by_pairs
raise KeyError((category_name, discriminator))
KeyError: ('renderer factories', '.xml')

Changing the view to:

@view_config(route_name='sitemap', renderer='static/sitemap.xml.jinja2')
def sitemap(request):
    ingredients = [ ingredient.name for ingredient in Cosmeceutical.get_all() ]
    products = [ product.name for product in Product.get_all() ]
    request.response.content_type = 'text/xml'
    return dict(ingredients=ingredients, products=products)

Gets past the KeyError from before, but gives me a 404 when I try navigating to mysite.com/static/sitemap.xml. What's going on?

EDIT: This is my sitemap.jinja2 file.

<?xml version="1.0" encoding="UTF-8"?>
<urlset
      xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
      xsi:schemaLocation="http://www.sitemaps.org/schemas/sitemap/0.9
            http://www.sitemaps.org/schemas/sitemap/0.9/sitemap.xsd">

{% for page in other_pages %}
<url>
  <loc>http://www.wisderm.com/{{page}}</loc>
  <changefreq>weekly</changefreq>
</url>
{% endfor %}

{% for ingredient in ingredients %}
<url>
  <loc>http://www.wisderm.com/ingredients/{{ingredient.replace(' ', '+')}}</loc>
  <changefreq>monthly</changefreq>
</url>
{% endfor %}

{% for product in products %}
<url>
  <loc>http://www.wisderm.com/products/{{product.replace(' ', '+')}}</loc>
  <changefreq>monthly</changefreq>
</url>
{% endfor %}

</urlset>

Answer 1

Given you want to establish http://example.com/sitemap.xml as your sitemap URL do that.

Add this line to init .py to register URL pattern http://example.com/sitemap.xml as route sitemap

config.add_route('sitemap', '/sitemap.xml')

register view code for route sitemap and render response with your custom jinja2 template sitemap.jinja2 . The file extension `jinja2' will trigger usage of jinja2 renderer.

@view_config(route_name='sitemap', renderer='static/sitemap.jinja2')
def sitemap(request):
    ingredients = [ ingredient.name for ingredient in Cosmeceutical.get_all() ]
    products = [ product.name for product in Product.get_all() ]
    return dict(ingredients=ingredients, products=products)

This will fix your errors resulting from trying to name your templates like URLs. But that mixed up renderer conventions shown below.

• *.pt triggers Chameleon renderer
• *.jinja2 triggers Jinja2 renderer
• *.mako triggers Mako renderer
• *.xml triggers XML renderer (that raised your first error)

Now it is still up to you to create XML based on sitemaps protocol . But your code looks promising. You pass your resource tree to the XML template. Every resource usually has access to their properties like URL or last_changed stuff.

Answer 2

Have a look here

from pyramid.threadlocal import get_current_registry

def mview(request):
    reg = get_current_registy

when I look into source there is method get_routes_mapper in pyramid.config.Configurator .

Maybe it helps You to generate sitemap on the 'fly' :)