如何在scrapy python中编写自定义链接提取器

我想编写我的自定义scrapy链接提取器来提取链接.

scrapy文档说它有两个内置的提取器.

http://doc.scrapy.org/en/latest/topics/link-extractors.html

但我还没有看到任何代码示例如何通过自定义链接提取器实现,有人可以给出一些编写自定义提取器的示例吗？

解决方法

class RCP_RegexLinkExtractor(SgmlLinkExtractor):
    """High performant link extractor"""

    def _extract_links(self,response_text,response_url,response_encoding,base_url=None):
        if base_url is None:
            base_url = urljoin(response_url,self.base_url) if self.base_url else response_url

        clean_url = lambda u: urljoin(base_url,remove_entities(clean_link(u.decode(response_encoding))))
        clean_text = lambda t: replace_escape_chars(remove_tags(t.decode(response_encoding))).strip()

        links_text = linkre.findall(response_text)
        urlstext = set([(clean_url(url),clean_text(text)) for url,_,text in links_text])

        return [Link(url,text) for url,text in urlstext]

用法

rules = (
    Rule(
        RCP_RegexLinkExtractor(
            allow=(r"epolls/2012/president/[a-z]{2}/[a-z]+_romney_vs_obama-[0-9]{4}.html"),# Regex explanation:
            #     [a-z]{2} - matches a two character state abbreviation
            #     [a-z]*   - matches a state name
            #     [0-9]{4} - matches a 4 number unique webpage identifier

            allow_domains=('realclearpolitics.com',),callback='parseStatePolls',# follow=None,# default 
        process_links='processLinks',process_request='processRequest',)

看看这里https://github.com/jtfairbank/RCP-Poll-Scraper

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