如何在Scrapy Python中創建規則？

我在創建規則時遇到了麻煩。讓想我的出發網址是http://www.example.com/search?q=news
當我打開這個網址的網頁瀏覽器我得到下面的源代碼：如何在Scrapy Python中創建規則？

<html><head>...</head><body> 
<ul id="results-list"> 
<li class="result clearfix news"> 
<div class="summary"> 
<h3><a href="/sports/hockey/struggling-canucks-rely-on-schneider-to-snag-win-against-sens/article2243069/">Struggling Canucks rely on Schneider to snag win against Sens</a></h3> 
<p class="summary">Nov 21, 2011&ndash; Eleventh place Canucks rely on goalie Cory Schneider to improve record to 10-9-1 
</p> 
<p class="meta"><a href="/sports/hockey/struggling-canucks-rely-on-schneider-to-snag-win-against-sens/article2243069/">http://www.example.com/sports/hockey/struggling-canucks-rely-on-schneider-to-snag-win-against-sens/article2243069/</a> 
</p> 
</div> 
</li> 
<li class="result clearfix news"> 
<div class="summary"> 
<h3><a href="/news/world/celebrities-set-to-testify-at-uk-media-ethics-inquiry/article2242840/">Celebrities set to testify at U.K. media ethics inquiry</a></h3> 
<p class="summary">Nov 20, 2011&ndash; Hugh Grant and J.K. Rowling given opportunity to strike back against tabloids’ invasion of privacy 
</p> 
<p class="meta"><a href="/news/world/celebrities-set-to-testify-at-uk-media-ethics-inquiry/article2242840/">http://www.example.com/news/world/celebrities-set-to-testify-at-uk-media-ethics-inquiry/article2242840/</a> 
</p> 
</div> 
</li> 
... 
</ul><!-- end of ul#results-list --> 

<ul class="paginator"> 
<li class="selected"><a href="http://www.example.com/search/?q=news&start=0">1</a></li> 
<li ><a href="http://www.example.com/search/?q=news&start=10">2</a></li> 
<li ><a href="http://www.example.com/search/?q=news&start=20">3</a></li> 
... 
<li class="jump last"><a href="http://www.example.com/search/?q=news&start=90">Last</a></li> 
</ul><!-- end of ul.paginator --> 
</body></html>

現在我想提取鏈接數據（該鏈接存在於UL＃結果列表）http://www.example.com/sports/hockey/struggling-canucks-rely-on-schneider-to-snag-win-against-sens/article2243069/等等...

我創建的蜘蛛爲如下：

from scrapy.contrib.spiders import CrawlSpider, Rule 
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor 
from scrapy.selector import HtmlXPathSelector 
from thirdapp.items import ThirdappItem 

class MySpider(CrawlSpider): 
    name = 'example.com' 
    allowed_domains = ['example.com'] 
    start_urls = [ 
     'http://www.example.com/search?q=news', 
     'http://www.example.com/search?q=movies', 
     ] 
    rules = (
     Rule(SgmlLinkExtractor(allow('?q=news',), restrict_xpaths('ul[@class="paginator"]',)), callback='parse_item', allow=True), 
     ) 

    def parse_item(self, response): 
     self.log('Hi, this is an item page! %s', response.url) 

     hxs = HtmlXPathSelector(response) 
     #item = ThirdappItem() 
     items = hxs.select('//h3') 
     scraped_items = [] 
     for item in items: 
      scraped_item = ThirdappItem() 
      scraped_item["title"] = item.select('a/text()').extract() 
      scraped_items.append(scraped_item) 
     return items 

spider = MySpider()

那麼會是什麼規則，使我會達到我期待的結果嗎？

來源

2011-11-21 Nilesh Tighare

首先，你期待什麼結果？秒，也許你應該處理你的規則中的鏈接，而不僅僅是包含列表項節點的ul-container，它具有所需的鏈接節點！

來源

2011-11-21 12:36:37

據文檔，SgmlLinkExtractor allow參數 - 單個正則表達式（或列表正則表達式），所述（絕對）網址必須以匹配要被提取。所以allow參數會是什麼樣子：

allow=('.*\?q=news.*',)

而且最有可能的，規則的最後一個參數是不是allow，但follow=True。

最終規則（注意，問號轉義字符）：

Rule(SgmlLinkExtractor(allow=('.*\?q=news.*',), restrict_xpaths=('ul[@class="paginator"]',)), callback='parse_item', follow=True)

來源

2011-11-22 18:31:14 reclosedev

如何在Scrapy Python中創建規則？

回答

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