我的JSON如何在每个结果之后返回NULL行？

Question

我正在尝试从公司注册簿中抓取一些数据，到目前为止，它可以抓取每个搜索结果，但是当我尝试导出它时。它会在每个搜索结果之后显示一个空对象，好像它会刮擦同一页两次？

这是日志的摘要。

2019-05-14 08:19:21 [scrapy.core.engine] DEBUG: Crawled (200) <GET https://www.companiesintheuk.co.uk/ltd/a-c-1> (referer: https://www.companiesintheuk.co.uk/Company/Find?q=a)
2019-05-14 08:19:21 [scrapy.core.scraper] DEBUG: Scraped from <200 https://www.companiesintheuk.co.uk/ltd/a-c-1>
{'location': u'BEANCROFT ROAD', 'postal_code': None, 'company_name': u'A C PLC', 'address': u'BEANCROFT FARM'}
2019-05-14 08:19:21 [scrapy.core.scraper] DEBUG: Scraped from <200 https://www.companiesintheuk.co.uk/ltd/a-c-1>
{'location': None, 'postal_code': None, 'company_name': None, 'address': None}

最后是我的代码

import scrapy
import re
from scrapy.linkextractors import LinkExtractor


class QuotesSpider(scrapy.Spider):

  name = 'CYRecursive'
  start_urls = [
      'https://www.companiesintheuk.co.uk/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,
      )

  def parse_details(self, response):

    # Looping throught the searchResult block and yielding it

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

提前谢谢！

Answer 1

您有两个元素div.col-md-6，每个公司页面一个（例如：https://www.companiesintheuk.co.uk/ltd/a-c-1）。因此，第一个包含公司详细信息，第二个包含地图而没有公司数据。

因此，您可以使用以下代码修改代码：

def parse_details(self, response):
    for i in response.css('div.col-md-6'):
        if not i.css('#content2 > strong:nth-child(2) > strong:nth-child(1)'):
            continue
        yield {
            'company_name': i.css('#content2 > strong:nth-child(2) > strong:nth-child(1) > div:nth-child(1)::text').get(),
            'address': i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > span:nth-child(1)::text").extract_first(),
            'location': i.css("#content2 > strong:nth-child(2) > address:nth-child(2) > div:nth-child(1) > span:nth-child(3)::text").extract_first(),
            'postal_code': 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(),
        }

因此，只需跳过最初不需要阻止的项目即可。