python - Scrapy回调函数

Question

我有一个基本的 scrapy 脚本，它执行以下操作:

1. 访问网站

2. 使用规则获取所有页面:

    rules = (
            Rule(LinkExtractor(allow=(), restrict_xpaths=('//*[@id="pagination_top"]/a',)), callback="parse_page", follow= True),
            )
   

3. 在每个页面中，获取产品页面的所有链接:

   def parse_page(self, response):
       for href in response.css("#prod_category > ul > li > a::attr('href')"):
           url = response.urljoin(href.extract())
           yield scrapy.Request(url, callback=self.parse_dir_contents)
   

4. 并访问每个产品页面以获取有关产品的详细信息。然后我从不同的链接获取更多详细信息

   def parse_dir_contents(self, response):
        # select xpath here
        print '________________________BEGIN PRODUCT________________________'
        item = detailedItem()
        item['title'] = sites.xpath('//*[@id="product-name"]/text()').extract()
   
        # get url_2 from this page
   
        request = scrapy.Request(url_2, callback=self.parse_detailed_contents)
        request.meta['item'] = item
        yield request
   

5. 最后是获取产品详细信息的函数

   I think this last parse_detailed_contents is where I have an issue

   def parse_detailed_contents(self, response):
       item = response.meta['item']
       sel = Selector(response)
       sites = sel.xpath('//*[@id="prod-details"]')
   
       print '________________________GETTING DETAILS________________________'
       item['prod_details'] = sites.xpath('//*[@id="prod-details"]/div/text()').extract()
   
       return item
   

问题是我的脚本为第一个链接返回 item['prod_details'] 但不为后续链接返回任何项目。

Is that because url_2 being passed in the same for all product?

有人可以帮忙吗？预先非常感谢!

Answer 1

尝试添加dont_filter=True

def parse_dir_contents(self, response):
 # select xpath here
 print '________________________BEGIN PRODUCT________________________'
 item = detailedItem()
 item['title'] = sites.xpath('//*[@id="product-name"]/text()').extract()

 # get url_2 from this page

 request = scrapy.Request(url_2, callback=self.parse_detailed_contents,dont_filter=True)
 request.meta['item'] = item
 yield request