Scrapy是否有可能从原始HTML数据中获取纯文本？

Question

例如：

scrapy shell http://scrapy.org/
content = hxs.select('//*[@id="content"]').extract()[0]
print content

然后，我得到以下原始HTML代码：

<div id="content">


  <h2>Welcome to Scrapy</h2>

  <h3>What is Scrapy?</h3>

  <p>Scrapy is a fast high-level screen scraping and web crawling
    framework, used to crawl websites and extract structured data from their
    pages. It can be used for a wide range of purposes, from data mining to
    monitoring and automated testing.</p>

  <h3>Features</h3>

  <dl>

    <dt>Simple</dt>
    <dt>
    </dt>
    <dd>Scrapy was designed with simplicity in mind, by providing the features
      you need without getting in your way
    </dd>

    <dt>Productive</dt>
    <dd>Just write the rules to extract the data from web pages and let Scrapy
      crawl the entire web site for you
    </dd>

    <dt>Fast</dt>
    <dd>Scrapy is used in production crawlers to completely scrape more than
      500 retailer sites daily, all in one server
    </dd>

    <dt>Extensible</dt>
    <dd>Scrapy was designed with extensibility in mind and so it provides
      several mechanisms to plug new code without having to touch the framework
      core

    </dd>
    <dt>Portable, open-source, 100% Python</dt>
    <dd>Scrapy is completely written in Python and runs on Linux, Windows, Mac and BSD</dd>

    <dt>Batteries included</dt>
    <dd>Scrapy comes with lots of functionality built in. Check <a
        href="http://doc.scrapy.org/en/latest/intro/overview.html#what-else">this
      section</a> of the documentation for a list of them.
    </dd>

    <dt>Well-documented &amp; well-tested</dt>
    <dd>Scrapy is <a href="/doc/">extensively documented</a> and has an comprehensive test suite
      with <a href="http://static.scrapy.org/coverage-report/">very good code
        coverage</a></dd>

    <dt><a href="/community">Healthy community</a></dt>
    <dd>
      1,500 watchers, 350 forks on Github (<a href="https://github.com/scrapy/scrapy">link</a>)<br>
      700 followers on Twitter (<a href="http://twitter.com/ScrapyProject">link</a>)<br>
      850 questions on StackOverflow (<a href="http://stackoverflow.com/tags/scrapy/info">link</a>)<br>
      200 messages per month on mailing list (<a
        href="https://groups.google.com/forum/?fromgroups#!aboutgroup/scrapy-users">link</a>)<br>
      40-50 users always connected to IRC channel (<a href="http://webchat.freenode.net/?channels=scrapy">link</a>)
    </dd>

    <dt><a href="/support">Commercial support</a></dt>
    <dd>A few companies provide Scrapy consulting and support</dd>

    <p>Still not sure if Scrapy is what you're looking for?. Check out <a
        href="http://doc.scrapy.org/en/latest/intro/overview.html">Scrapy at a
      glance</a>.

    </p>
    <h3>Companies using Scrapy</h3>

    <p>Scrapy is being used in large production environments, to crawl
      thousands of sites daily. Here is a list of <a href="/companies/">Companies
        using Scrapy</a>.</p>

    <h3>Where to start?</h3>

    <p>Start by reading <a href="http://doc.scrapy.org/en/latest/intro/overview.html">Scrapy at a glance</a>,
      then <a href="/download/">download Scrapy</a> and follow the <a
          href="http://doc.scrapy.org/en/latest/intro/tutorial.html">Tutorial</a>.


    </p></dl>
</div>

但是我想直接从scrapy 获取纯文本。

我不希望使用任何XPath选择提取p，h2，h3…标签，因为我爬一个网站，其主要内容嵌入到table，tbody; 递归地 找到xPath可能是一项繁琐的任务。

可以通过Scrapy中的内置函数来实现吗？还是我需要外部工具进行转换？我已经阅读了Scrapy的所有文档，但一无所获。

Answer 1

Scrapy没有内置此类功能。html2text是你要寻找的。

这是一个示例spider，它抓取Wikipedia的python页面，使用xpath获取第一段，然后使用以下命令将html转换为纯文本html2text：

from scrapy.selector import HtmlXPathSelector
from scrapy.spider import BaseSpider
import html2text


class WikiSpider(BaseSpider):
    name = "wiki_spider"
    allowed_domains = ["www.wikipedia.org"]
    start_urls = ["http://en.wikipedia.org/wiki/Python_(programming_language)"]

    def parse(self, response):
        hxs = HtmlXPathSelector(response)
        sample = hxs.select("//div[@id='mw-content-text']/p[1]").extract()[0]

        converter = html2text.HTML2Text()
        converter.ignore_links = True
        print(converter.handle(sample)) #Python 3 print syntax

印刷品：

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