Question

我一直在尝试从公司注册簿中获取一些信息。哪个可行，但我想对搜索条目给出的每个结果重复一遍。我一直在尝试使用linkextractors，但即时通讯无法正常工作。

搜索结果网页为： https://www.companiesintheuk.co.uk/Company/Find?q=a

从搜索项中抓取单个结果是可行的（如果我单击一个结果项），但是如何对每个结果项重复此操作？

这是我的代码：

import scrapy
import re
from scrapy.linkextractors import LinkExtractor

class QuotesSpider(scrapy.Spider):


  name = 'CYRecursive'
  start_urls = [
      'https://www.companiesintheuk.co.uk/ltd/a-2']

  def parse(self, response):

    # Looping throught the searchResult block and yielding it
    for i in response.css('div.col-md-9'):

        for i in response.css('div.col-md-6'):
          yield {
              'company_name': re.sub('\s+', ' ', ''.join(i.css('#content2 > strong:nth-child(2) > strong:nth-child(1) > div:nth-child(1)::text').get())),
              'address': re.sub('\s+', ' ', ''.join(i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > span:nth-child(1)::text").extract_first())),
              'location': re.sub('\s+', ' ', ''.join(i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > span:nth-child(3)::text").extract_first())),
              'postal_code': re.sub('\s+', ' ', ''.join(i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > a:nth-child(5) > span:nth-child(1)::text").extract_first())),
          }

Answer 1

import scrapy
import re
from scrapy.linkextractors import LinkExtractor


class QuotesSpider(scrapy.Spider):

    name = 'CYRecursive'
    start_urls = [
        'https://www.companiesintheuk.co.uk/Company/Find?q=a']

    def parse(self, response):

        for company_url in response.xpath('//div[@class="search_result_title"]/a/@href').extract():
            yield scrapy.Request(
                url=response.urljoin(company_url),
                callback=self.parse_details,
            )

        next_page_url = response.xpath('//li/a[@class="pageNavNextLabel"]/@href').extract_first()
        if next_page_url:
            yield scrapy.Request(
                url=response.urljoin(next_page_url),
                callback=self.parse,
            )


    def parse_details(self, response):

        # Looping throught the searchResult block and yielding it
        for i in response.css('div.col-md-9'):

            for i in response.css('div.col-md-6'):
                yield {
                    'company_name': re.sub('\s+', ' ', ''.join(i.css('#content2 > strong:nth-child(2) > strong:nth-child(1) > div:nth-child(1)::text').get())),
                    'address': re.sub('\s+', ' ', ''.join(i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > span:nth-child(1)::text").extract_first())),
                    'location': re.sub('\s+', ' ', ''.join(i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > span:nth-child(3)::text").extract_first())),
                    'postal_code': re.sub('\s+', ' ', ''.join(i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > a:nth-child(5) > span:nth-child(1)::text").extract_first())),
                }

当然，您可以使用start_requests自动yield进行从a到z的所有搜索。

您的CSS表达式错误：

            yield {
                'company_name': response.xpath('//div[@itemprop="name"]/text()').extract_first(),
                'address': response.xpath('//span[@itemprop="streetAddress"]/text()').extract_first(),
                'location': response.xpath('//span[@itemprop="addressLocality"]/text()').extract_first(),
                'postal_code': response.xpath('//span[@itemprop="postalCode"]/text()').extract_first(),
            }

如何抓取每个搜索项的结果并返回？

1 个答案: