Question

我正在尝试使用scrapy抓取this网站。页面结构如下所示：

<div class="list">
  <a id="follows" name="follows"></a>
 <h4 class="li_group">Follows</h4>
 <div class="soda odd"><a href="...">Star Trek</a></div> 
 <div class="soda even"><a href="...</a></div>
 <div class="soda odd"><a href="..">Star Trek: The Motion Picture</a></div>
 <div class="soda even"><a href="..">Star Trek II: The Wrath of Khan</a></div>
 <div class="soda odd"><a href="..">Star Trek III: The Search for Spock</a></div>
 <div class="soda even"><a href="..">Star Trek IV: The Voyage Home</a></div>
  <a id="followed_by" name="followed_by"></a>
 <h4 class="li_group">Followed by</h4>
 <div class="soda odd"><a href="..">Star Trek V: The Final Frontier</a></div>
 <div class="soda even"><a href="..">Star Trek VI: The Undiscovered Country</a></div>
 <div class="soda odd"><a href="..">Star Trek: Deep Space Nine</a></div>
 <div class="soda even"><a href="..">Star Trek: Generations</a></div>
 <div class="soda odd"><a href="..">Star Trek: Voyager</a></div>
 <div class="soda even"><a href="..">First Contact</a></div>
   <a id="spin_off" name="spin_off"></a>
 <h4 class="li_group">Spin-off</h4>
 <div class="soda odd"><a href="..">Star Trek: The Next Generation - The Transinium Challenge</a></div>
 <div class="soda even"><a href="..">A Night with Troi</a></div>
 <div class="soda odd"><a href="..">Star Trek: Deep Space Nine</a></div
</div>

我想在<h4 class="li_group">Follows</h4>和<h4 class="li_group">Followed by</h4>之间选择和提取文字，然后在<h4 class="li_group">Followed by</h4>和<h4 class="li_group">Spin-off</h4>之间提交文字。我试过这段代码：

def parse(self, response):
    for sel in response.css("div.list"):
        item = ImdbcoItem()
        item['Follows'] = sel.css("a#follows+h4.li_group ~ div a::text").extract(),
        item['Followed_by'] = sel.css("a#vfollowed_by+h4.li_group ~ div a::text").extract(),
        item['Spin_off'] = sel.css("a#spin_off+h4.li_group ~ div a::text").extract(),
    return item

但是，第一项提取所有div不仅仅是<h4 class="li_group">Follows</h4>和<h4 class="li_group">Followed by</h4>之间的div 任何帮助都会对您有所帮助!!

Answer 1

您可以尝试使用以下XPath表达式来获取

“Follows”块的所有文本节点：

//div[./preceding-sibling::h4[1]="Follows"]//text()

“跟随”块的所有文本节点：

//div[./preceding-sibling::h4[1]="Followed by"]//text()

“Spin off”块的所有文本节点：

//div[./preceding-sibling::h4[1]="Spin-off"]//text()

Answer 2

我喜欢用于这些案例的提取模式是：

遍历“边界”（此处为h4元素）
从1
使用XPath的following-sibling轴，就像@ Andersson的回答一样，在下一个边界之前获取元素，
并通过计算前面“边界”元素的数量来过滤它们，因为我们从枚举中知道我们在哪里

这将是循环：

$ scrapy shell 'http://www.imdb.com/title/tt0092455/trivia?tab=mc&ref_=tt_trv_cnn'
(...)
>>> for cnt, h4 in enumerate(response.css('div.list > h4.li_group'), start=1):
...     print(cnt, h4.xpath('normalize-space()').get())
... 
1 Follows 
2 Followed by 
3 Edited into 
4 Spun-off from 
5 Spin-off 
6 Referenced in 
7 Featured in 
8 Spoofed in

这是使用枚举来获取边界之间元素的一个示例（请注意，这在表达式中使用带有$cnt的XPath变量并在cnt=cnt中传递.xpath()）：

>>> for cnt, h4 in enumerate(response.css('div.list > h4.li_group'), start=1):
...     print(cnt, h4.xpath('normalize-space()').get())
...     print(h4.xpath('following-sibling::div[count(preceding-sibling::h4)=$cnt]',
                       cnt=cnt).xpath(
                          'string(.//a)').getall())
... 
1 Follows 
['Star Trek', 'Star Trek: The Animated Series', 'Star Trek: The Motion Picture', 'Star Trek II: The Wrath of Khan', 'Star Trek III: The Search for Spock', 'Star Trek IV: The Voyage Home']
2 Followed by 
['Star Trek V: The Final Frontier', 'Star Trek VI: The Undiscovered Country', 'Star Trek: Deep Space Nine', 'Star Trek: Generations', 'Star Trek: Voyager', 'First Contact', 'Star Trek: Insurrection', 'Star Trek: Enterprise', 'Star Trek: Nemesis', 'Star Trek', 'Star Trek Into Darkness', 'Star Trek Beyond', 'Star Trek: Discovery', 'Untitled Star Trek Sequel']
3 Edited into 
['Reading Rainbow: The Bionic Bunny Show', 'The Unauthorized Hagiography of Vincent Price']
4 Spun-off from 
['Star Trek']
5 Spin-off 
['Star Trek: The Next Generation - The Transinium Challenge', 'A Night with Troi', 'Star Trek: Deep Space Nine', "Star Trek: The Next Generation - Future's Past", 'Star Trek: The Next Generation - A Final Unity', 'Star Trek: The Next Generation: Interactive VCR Board Game - A Klingon Challenge', 'Star Trek: Borg', 'Star Trek: Klingon', 'Star Trek: The Experience - The Klingon Encounter']
6 Referenced in 
(...)

以下是如何使用它来填充和项目（这里，我使用简单的dict只是为了说明）：

>>> item = {}
>>> for cnt, h4 in enumerate(response.css('div.list > h4.li_group'), start=1):
...     key = h4.xpath('normalize-space()').get().strip() # there are some non-breaking spaces
...     if key in ['Follows', 'Followed by', 'Spin-off']:
...         values = h4.xpath('following-sibling::div[count(preceding-sibling::h4)=$cnt]',
...                        cnt=cnt).xpath(
...                           'string(.//a)').getall()
...         item[key] = values
... 

>>> from pprint import pprint
>>> pprint(item)
{'Followed by': ['Star Trek V: The Final Frontier',
                 'Star Trek VI: The Undiscovered Country',
                 'Star Trek: Deep Space Nine',
                 'Star Trek: Generations',
                 'Star Trek: Voyager',
                 'First Contact',
                 'Star Trek: Insurrection',
                 'Star Trek: Enterprise',
                 'Star Trek: Nemesis',
                 'Star Trek',
                 'Star Trek Into Darkness',
                 'Star Trek Beyond',
                 'Star Trek: Discovery',
                 'Untitled Star Trek Sequel'],
 'Follows': ['Star Trek',
             'Star Trek: The Animated Series',
             'Star Trek: The Motion Picture',
             'Star Trek II: The Wrath of Khan',
             'Star Trek III: The Search for Spock',
             'Star Trek IV: The Voyage Home'],
 'Spin-off': ['Star Trek: The Next Generation - The Transinium Challenge',
              'A Night with Troi',
              'Star Trek: Deep Space Nine',
              "Star Trek: The Next Generation - Future's Past",
              'Star Trek: The Next Generation - A Final Unity',
              'Star Trek: The Next Generation: Interactive VCR Board Game - A '
              'Klingon Challenge',
              'Star Trek: Borg',
              'Star Trek: Klingon',
              'Star Trek: The Experience - The Klingon Encounter']}
>>>

如何在两个元素之间选择和提取文本？

2 个答案: