如何正确使用规则,restrict_xpaths使用scrapy来抓取和解析URL？

Question

我正在尝试编写爬行蜘蛛来抓取网站的RSS源,然后解析文章的元标记.

第一个RSS页面是显示RSS类别的页面.我设法提取链接,因为标记位于标记中.它看起来像这样:

        <tr>
           <td class="xmlLink">
             <a href="http://feeds.example.com/subject1">subject1</a>
           </td>   
        </tr>
        <tr>
           <td class="xmlLink">
             <a href="http://feeds.example.com/subject2">subject2</a>
           </td>
        </tr>

单击该链接后,它会为您显示该RSS类别的文章,如下所示:

   <li class="regularitem">
    <h4 class="itemtitle">
        <a href="http://example.com/article1">article1</a>
    </h4>
  </li>
  <li class="regularitem">
     <h4 class="itemtitle">
        <a href="http://example.com/article2">article2</a>
     </h4>
  </li>

正如您所看到的,如果我使用标记我可以再次获取xpath的链接,我希望我的抓取工具转到该标记内的链接并为我解析元标记.

这是我的抓取代码:

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from tutorial.items import exampleItem


class MetaCrawl(CrawlSpider):
    name = 'metaspider'
    start_urls = ['http://example.com/tools/rss'] # urls from which the spider will start crawling
    rules = [Rule(SgmlLinkExtractor(restrict_xpaths=('//td[@class="xmlLink"]')), follow=True),
        Rule(SgmlLinkExtractor(restrict_xpaths=('//h4[@class="itemtitle"]')), callback='parse_articles')]

    def parse_articles(self, response):
        hxs = HtmlXPathSelector(response)
        meta = hxs.select('//meta')
        items = []
        for m in meta:
           item = exampleItem()
           item['link'] = response.url
           item['meta_name'] =m.select('@name').extract()
           item['meta_value'] = m.select('@content').extract()
           items.append(item)
        return items

但是,这是我运行爬虫时的输出:

DEBUG: Crawled (200) <GET http://http://feeds.example.com/subject1> (referer: http://example.com/tools/rss)
DEBUG: Crawled (200) <GET http://http://feeds.example.com/subject2> (referer: http://example.com/tools/rss)

我在这做错了什么？我一遍又一遍地阅读文档,但我觉得我一直在忽略一些东西.任何帮助,将不胜感激.

编辑:添加:items.append(项目).在原帖中忘记了它. 编辑::我也试过这个,结果输出相同:

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from reuters.items import exampleItem
from scrapy.http import Request

class MetaCrawl(CrawlSpider):
    name = 'metaspider'
    start_urls = ['http://example.com/tools/rss'] # urls from which the spider will start crawling
    rules = [Rule(SgmlLinkExtractor(allow=[r'.*',], restrict_xpaths=('//td[@class="xmlLink"]')), follow=True),
             Rule(SgmlLinkExtractor(allow=[r'.*'], restrict_xpaths=('//h4[@class="itemtitle"]')),follow=True),]


    def parse(self, response):       
        hxs = HtmlXPathSelector(response)
        meta = hxs.select('//td[@class="xmlLink"]/a/@href')
        for m in meta:
            yield Request(m.extract(), callback = self.parse_link)


    def parse_link(self, response):       
        hxs = HtmlXPathSelector(response)
        meta = hxs.select('//h4[@class="itemtitle"]/a/@href')
        for m in meta:
            yield Request(m.extract(), callback = self.parse_again)    

    def parse_again(self, response):
        hxs = HtmlXPathSelector(response)
        meta = hxs.select('//meta')
        items = []
        for m in meta:
            item = exampleItem()
            item['link'] = response.url
            item['meta_name'] = m.select('@name').extract()
            item['meta_value'] = m.select('@content').extract()
            items.append(item)
        return items

Answer 1

您返回了一个空的items，您需要追加item到items.
也可以yield item在循环中。