Question

我是Scrapy的新手，我正在尝试抓取page并从商品中获取价格，问题是scrapy正在返回无序的值，我不知道为什么。

这是我的简单代码

import scrapy
from ..items import AmazonItem
from scrapy.http import Request
import time

class QuotesSpider(scrapy.Spider):
    name = "main"

    def start_requests(self):
        urls = [
            'https://www.amazon.com/best-sellers-movies-TV-DVD-Blu-ray/zgbs/movies-tv/ref=zg_bs_nav_0',
        ]
        for url in urls:
            yield scrapy.Request(url=url, callback=self.parse)

    def parse(self, response):

        # amazon = AmazonItem()

        ol_response = response.xpath('//ol[@id="zg-ordered-list"]/li')

        for number_ra in range(0,50):
             response_div = ol_response[number_ra]
             price = response_div.css(".p13n-sc-price::text").extract()
             item_name = response_div.xpath("span/div/span/a/div/text()").get().strip()
             link = response_div.xpath("span/div/span/a").attrib['href'].split('/')[3].split('?')[0]
             print("({}) {} , PRICE: {}".format(number_ra+1,item_name,price))
             print(link+"\n")

名称和ID的顺序正确，但价格不正确。

谢谢，伙计们

Answer 1

您做错了方式

您应该一个一个地遍历每个项目

def解析（自身，响应）：

for item in response.xpath('//ol[@id="zg-ordered-list"]/li'):
     price = item.css(".p13n-sc-price::text").get()
     item_name = item.css(".p13n-sc-truncate.p13n-sc-line-clamp-1::text").get()
     link = response.urljoin(item.css(".a-link-normal::attr(href)").get())
     print("{} , PRICE: {}".format(item_name,price))

抓取时抓取返回无序值

1 个答案: