python - scrapy如何正确使用Rules、restrict_xpaths抓取解析URL？

Question

我正在尝试编写一个抓取蜘蛛程序来抓取网站的 RSS 提要，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。我设法提取了链接，因为标签在标签中。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，它会为您带来该 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>

如您所见，如果我使用标签，我可以再次获得 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。

这是我的爬虫代码:

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

然而，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)

我在这里做错了什么？我一直在一遍又一遍地阅读文档，但我觉得我总是忽略了一些东西。任何帮助将不胜感激。

编辑: 添加:items.append(item) 。原来的帖子忘记了。编辑::我也试过了，结果是一样的输出:

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，您需要将item附加到items。
您还可以在循环中yield item。