xpath - 如何使用hxs.select完整地获取整个文档

我已经在这里呆了12个小时，希望有人能帮上忙。

这是我的代码，我想要的是在页面爬行时获得页面上每个链接的锚点和URL。

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from scrapy.utils.url import urljoin_rfc
from scrapy.utils.response import get_base_url
from urlparse import urljoin

#from scrapy.item import Item
from tutorial.items import DmozItem

class HopitaloneSpider(CrawlSpider):
name = 'dmoz'
allowed_domains = ['domain.co.uk']
start_urls = [
    'http://www.domain.co.uk'
]

rules = (
    #Rule(SgmlLinkExtractor(allow='>example\.org', )),
    Rule(SgmlLinkExtractor(allow=('\w+$', )), callback='parse_item', follow=True),
)

user_agent = 'Mozilla/5.0 (Windows; U; MSIE 9.0; WIndows NT 9.0; en-US))'

def parse_item(self, response):
    #self.log('Hi, this is an item page! %s' % response.url)

    hxs = HtmlXPathSelector(response)
    #print response.url
    sites = hxs.select('//html')
    #item = DmozItem()
    items = []

    for site in sites:

                   item = DmozItem()
                   item['title'] = site.select('a/text()').extract()
                   item['link'] = site.select('a/@href').extract()

                   items.append(item)

    return items

我做错了...我的眼睛现在疼。

最佳答案

response.body应该是你想要的

def parse_item(self, response):
    #self.log('Hi, this is an item page! %s' % response.url)

    body = response.body
    item = ....

关于xpath - 如何使用hxs.select完整地获取整个文档，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/13435620/