通过scrapy抓取网站的sitemap信息

# This script allows you to quickly crawl pages based off of the sitemap(s) for any given domain.  The XmlXPathSelector was not doing it for me nor was the XmlItemExporter this code is by far the fastest and easiest way I have found to crawl a site based off of the URLs listed in the sitemap.
import re
from scrapy.spider import BaseSpider
from scrapy import log
from scrapy.utils.response import body_or_str
from scrapy.http import Request
from scrapy.selector import HtmlXPathSelector
class SitemapSpider(BaseSpider):
    name = "SitemapSpider"
    start_urls = ["http://www.domain.com/sitemap.xml"]
    def parse(self, response):
        nodename = 'loc'
        text = body_or_str(response)
        r = re.compile(r"(<%s[\s>])(.*?)(</%s>)" % (nodename, nodename), re.DOTALL)
        for match in r.finditer(text):
            url = match.group(2)
            yield Request(url, callback=self.parse_page)
    def parse_page(self, response):
                hxs = HtmlXPathSelector(response)
                #Mock Item
        blah = Item()
        #Do all your page parsing and selecting the elemtents you want
                blash.divText = hxs.select('//div/text()').extract()[0]
        yield blah