Scrapy能否从HTML原始数据中获取纯文本？

Question

例如：

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

然后，我得到了以下原始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中获取纯文本内容。

我不想使用任何XPath选择器来提取p，h2，h3 ...标签，因为我正在爬取一个将主要内容嵌入到table，tbody中的网站；而且还具有递归结构。 找到XPath可能会很繁琐。

这可以通过Scrapy中的内置功能实现吗？还是需要外部工具进行转换？我已经阅读了所有Scrapy的文档，但没有得到任何帮助。

这是一个可以将原始HTML转换为纯文本的示例网站：http://beaker.mailchimp.com/html-to-text

- inix

---

使用内置类或函数？ - inix

Answer 1

Scrapy并没有内置这样的功能。你需要使用html2text。

这是一个示例spider，它会爬取维基百科Python页面，使用xpath获取第一段，并使用html2text将HTML转换为纯文本：

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

输出：

**Python**是一种广泛使用的通用高级编程语言。[11][12][13] 它的设计哲学强调代码可读性， 并且它的语法允许程序员用比如C等语言更少的代码行表达概念。[14][15] 该语言提供了旨在实现小型和大型清晰程序的结构。[16]

- alecxe

Answer 2

使用 lxml.html 的 tostring() 方法，参数为 method="text"，是另一种解决方法。在 Scrapy 内部使用了 lxml。通常你需要的参数是 encoding=unicode。

有关详细信息，请参见http://lxml.de/api/lxml.html-module.html。

from scrapy.spider import BaseSpider
import lxml.etree
import lxml.html

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):
        root = lxml.html.fromstring(response.body)

        # optionally remove tags that are not usually rendered in browsers
        # javascript, HTML/HEAD, comments, add the tag names you dont want at the end
        lxml.etree.strip_elements(root, lxml.etree.Comment, "script", "head")

        # complete text
        print lxml.html.tostring(root, method="text", encoding=unicode)

        # or same as in alecxe's example spider,
        # pinpoint a part of the document using XPath
        #for p in root.xpath("//div[@id='mw-content-text']/p[1]"):
        #   print lxml.html.tostring(p, method="text")

- paul trmbrth

---

我已经阅读了lxml文档，它确实是一个强大的工具，非常感谢。 - inix

不用谢。确实，lxml可以胜任所有任务，并且学起来也不难。 - paul trmbrth

谢谢您，先生。这对我很有帮助。 - B.T Gutsu

Answer 3

此时，我认为您不需要安装任何第三方库。Scrapy使用选择器提供此功能：
假设这是一个复杂的选择器：

sel = Selector(text='<a href="#">Click here to go to the <strong>Next Page</strong></a>')

我们可以使用以下方法获取整个文本：

text_content = sel.xpath("//a[1]//text()").extract()
# which results [u'Click here to go to the ', u'Next Page']

然后你可以轻松地将它们拼合在一起：

   ' '.join(text_content)
   # Click here to go to the Next Page

- Reyraa