3

我使用scrapy来抓取一个网站。

我想提取某些 div 的内容。

<div class="short-description">
{some mess with text, <br>, other html tags, etc}
</div>

loader.add_xpath('short_description', "//div[@class='short-description']/div")

通过该代码，我得到了我需要的东西，但结果包括包装 html ( <div class="short-description">...</div>)

如何摆脱那个父html标签？

注意。像 text()、node() 这样的选择器无法帮助我，因为我的 div 包含<br>, <p>, other divs, etc.、空格，我需要保留它们。

4

2 回答 2

2

尝试node()结合Join()：

loader.get_xpath('//div[@class="short-description"]/node()', Join())

结果看起来像：

>>> from scrapy.contrib.loader import XPathItemLoader
>>> from scrapy.contrib.loader.processor import Join
>>> from scrapy.http import HtmlResponse
>>>
>>> body = """
...     <html>
...         <div class="short-description">
...             {some mess with text, <br>, other html tags, etc}
...             <div>
...                 <p>{some mess with text, <br>, other html tags, etc}</p>
...             </div>
...             <p>{some mess with text, <br>, other html tags, etc}</p>
...         </div>
...     </html>
... """
>>> response = HtmlResponse(url='http://example.com/', body=body)
>>>
>>> loader = XPathItemLoader(response=response)
>>>
>>> print loader.get_xpath('//div[@class="short-description"]/node()', Join())

            {some mess with text,  <br> , other html tags, etc}
             <div>
                <p>{some mess with text, <br>, other html tags, etc}</p>
            </div>
             <p>{some mess with text, <br>, other html tags, etc}</p>
>>>
>>> loader.get_xpath('//div[@class="short-description"]/node()', Join())
u'\n            {some mess with text,  <br> , other html tags, etc}\n
   <div>\n         <p>{some mess with text, <br>, other html tags, etc}</p>\n
   </div> \n     <p>{some mess with text, <br>, other html tags, etc}</p> \n'
于 2013-03-26T03:33:00.660 回答
2 
hxs = HtmlXPathSelector(response)
for text in hxs.select("//div[@class='short-description']/text()").extract(): 
    print text
于 2013-03-26T01:35:55.603 回答