Question

我创建了一个扩展CrawlSpider的蜘蛛，并遵循http://scrapy.readthedocs.org/en/latest/topics/spiders.html的建议

问题是我需要解析起始网址（恰好与主机名一致）和它所包含的一些链接。

所以我已经定义了一条规则：rules = [Rule(SgmlLinkExtractor(allow=['/page/d+']), callback='parse_items', follow=True)]，但没有任何反应。

然后我尝试定义一组规则，例如：rules = [Rule(SgmlLinkExtractor(allow=['/page/d+']), callback='parse_items', follow=True), Rule(SgmlLinkExtractor(allow=['/']), callback='parse_items', follow=True)]。现在的问题是蜘蛛解析一切。

如何告诉蜘蛛解析_start_url_以及它包含的一些链接？

更新

我试图覆盖parse_start_url方法，所以现在我可以从起始页面获取数据，但它仍然不会跟随使用Rule定义的链接：

class ExampleSpider(CrawlSpider):
  name = 'TechCrunchCrawler'
  start_urls = ['http://techcrunch.com']
  allowed_domains = ['techcrunch.com']
  rules = [Rule(SgmlLinkExtractor(allow=['/page/d+']), callback='parse_links', follow=True)]

  def parse_start_url(self, response):
      print '++++++++++++++++++++++++parse start url++++++++++++++++++++++++'
      return self.parse_links(response)

  def parse_links(self, response):
      print '++++++++++++++++++++++++parse link called++++++++++++++++++++++++'
      articles = []
      for i in HtmlXPathSelector(response).select('//h2[@class="headline"]/a'):
          article = Article()
          article['title'] = i.select('./@title').extract()
          article['link'] = i.select('./@href').extract()
          articles.append(article)

      return articles

Answer 1

我过去也遇到过类似的问题我坚持使用BaseSpider。

试试这个：

from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector
from scrapy.http import Request
from scrapy.contrib.loader import XPathItemLoader

from techCrunch.items import Article


class techCrunch(BaseSpider):
    name = 'techCrunchCrawler'
    allowed_domains = ['techcrunch.com']

    # This gets your start page and directs it to get parse manager
    def start_requests(self):
        return [Request("http://techcrunch.com", callback=self.parseMgr)]

    # the parse manager deals out what to parse and start page extraction
    def parseMgr(self, response):
        print '++++++++++++++++++++++++parse start url++++++++++++++++++++++++'
        yield self.pageParser(response)

        nextPage = HtmlXPathSelector(response).select("//div[@class='page-next']/a/@href").extract()
        if nextPage:
            yield Request(nextPage[0], callback=self.parseMgr)

    # The page parser only parses the pages and returns items on each page call
    def pageParser(self, response):
        print '++++++++++++++++++++++++parse link called++++++++++++++++++++++++'
        loader = XPathItemLoader(item=Article(), response=response)
        loader.add_xpath('title', '//h2[@class="headline"]/a/@title')
        loader.add_xpath('link', '//h2[@class="headline"]/a/@href')
        return loader.load_item()

无法使用Scrapy关注链接

1 个答案: