无需报废即可跳转到下一页

Question

我正在尝试跳转到下一页而不会报废，并在下一页开始报废。

这是我的代码。

import scrapy
import re


class ProductSpider(scrapy.Spider):
    name = 'product'
    start_url = ['https://www.google.nl/search?client=opera&biw=1880&bih=1008&output=search&tbm=shop&q=738678181690&oq=738678181690&gs_l=products-cc.12...0.0.0.2438282.0.0.0.0.0.0.0.0..0.0....0...1ac..64.products-cc..0.0.0....0.65SJMExjNxE#spd=0']
    start_url = response.css('.MCpGKc > a::attr(href)').extract_first()
    if start_url:
        start_url = response.urljoin(start_url)
        yield scrapy.Request(url=next_page_url, callback=self.parse)

这里我开始报废，问题就在上面。

    def parse(self, response):
        self.log('Bla Bla Bla:' + response.url)
        for product in response.css(".os-main-table"):
            item ={
            "all_sellers": product.css(".os-seller-name-primary > a::text").extract(),
            "all_prices": product.css("td.os-total-col::text").re("\d+\,\d{1,2}"),
            }
        for item in zip(all_prices,all_sellers):
            scrapped_info = {
                'all_sellers' : item[0],
                'all_prices': item[1],
                }
            yield scrapped_info


        next_page_url =  response.css('.pag-prev-next-links > a:last-child::attr(href)').extract_first()
        if next_page_url:
            next_page_url = response.urljoin(next_page_url)
            yield scrapy.Request(url=next_page_url, callback=self.parse)