Scrapy忽略noindex

Question

我正在抓取大量的网址,并想知道是否有可能让scrapy不解析带有'meta name ="robots"content ="noindex"'的网页？查看此处列出的拒绝规则http://doc.scrapy.org/en/latest/topics/link-extractors.html,看起来deny规则仅适用于URL.你有scrapy忽略xpath吗？

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

from wallspider.items import Website


class Spider(CrawlSpider):
    name = "browsetest"
    allowed_domains = ["www.mydomain.com"]
    start_urls = ["http://www.mydomain.com",]

    rules = (
        Rule(SgmlLinkExtractor(allow=('/browse/')), callback="parse_items", follow= True),
        Rule(SgmlLinkExtractor(allow=(),unique=True,deny=('/[1-9]$', '(bti=)[1-9]+(?:\.[1-9]*)?', '(sort_by=)[a-zA-Z]', '(sort_by=)[1-9]+(?:\.[1-9]*)?', '(ic=32_)[1-9]+(?:\.[1-9]*)?', '(ic=60_)[0-9]+(?:\.[0-9]*)?', '(search_sort=)[1-9]+(?:\.[1-9]*)?', 'browse-ng.do\?', '/page/', '/ip/', 'out\+value', 'fn=', 'customer_rating', 'special_offers', 'search_sort=&', 'facet=' ))),
    )

    def parse_items(self, response):
        hxs = HtmlXPathSelector(response)
        sites = hxs.select('//html')
        items = []

        for site in sites:
            item = Website()
            item['url'] = response.url
            item['canonical'] = site.xpath('//head/link[@rel="canonical"]/@href').extract()
            item['robots'] = site.select('//meta[@name="robots"]/@content').extract()
            items.append(item)

        return items

Answer 1

不幸的是,CrawlSpider它没有提供您想要做的选项.不过,您可以覆盖其方法来实现这一点.

尝试将此方法添加到您的蜘蛛:

    def _response_downloaded(self, response):
        # Check whether this page contains the meta noindex in order to skip the processing.
        sel = Selector(response)
        if sel.xpath('//meta[@content="noindex"]'):
            return

        return super(Spider, self)._response_downloaded(response)

每当文档不够时,您可以检查源代码以查看可以更改的内容以及位置,只需注意您使用的是哪个版本.您可以在github中浏览最新的源代码:https://github.com/scrapy/scrapy/blob/master/scrapy/contrib/spiders/crawl.py#L61

但最好检查系统中的源代码.如果您正在使用可以轻松完成??操作员的IPython .