python之如何使用 Scrapy 抓取新链接

我最近才开始用Scrapy，所以对它不是很熟练，所以这真的是一个新手问题。

我正在抓取一些随机惯例进行练习，我抓取了名称和展位号，但我还想要来自公司的链接，这些链接位于新窗口内，我已经找到并存储了来自 anchor 标签的链接，但是我不知道如何抓取这些新链接，任何形式的帮助或指导都将非常有用

import scrapy 
 
class ConventionSpider(scrapy.Spider): 
    name = 'convention' 
    allowed_domains = ['events.jspargo.com/ASCB18/Public/Exhibitors.aspx?sortMenu=102003'] 
    start_urls = ['https://events.jspargo.com/ASCB18/Public/Exhibitors.aspx?sortMenu=102003'] 
 
    def parse(self, response): 
        name = response.xpath('//*[@class="companyName"]') 
        number = response.xpath('//*[@class="boothLabel"]') 
        link = response.xpath('//*[@class="companyName"]') 
        for row, row1, row2 in zip(name, number, link): 
            company = row.xpath('.//*[@class="exhibitorName"]/text()').extract_first() 
            booth_num = row1.xpath('.//*[@class="boothLabel aa-mapIt"]/text()').extract_first() 
            url = row2.xpath('.//a/@href').extract_first() 
 
            yield {'Company': company,'Booth Number': booth_num} 

请您参考如下方法:

参见此以供引用 https://github.com/NilanshBansal/Craigslist_Scrapy/blob/master/craigslist/spiders/jobs.py

import scrapy 
from scrapy import Request 
 
class ConventionSpider(scrapy.Spider): 
name = 'convention' 
# allowed_domains = ['events.jspargo.com/ASCB18/Public/Exhibitors.aspx?sortMenu=102003'] 
start_urls = ['https://events.jspargo.com/ASCB18/Public/Exhibitors.aspx?sortMenu=102003'] 
 
def parse(self, response): 
    name = response.xpath('//*[@class="companyName"]') 
    number = response.xpath('//*[@class="boothLabel"]') 
    link = response.xpath('//*[@class="companyName"]') 
    for row, row1, row2 in zip(name, number, link): 
        company = row.xpath('.//*[@class="exhibitorName"]/text()').extract_first() 
        booth_num = row1.xpath('.//*[@class="boothLabel aa-mapIt"]/text()').extract_first() 
        url = row2.xpath('.//a/@href').extract_first() 
 
        yield Request(url,callback=self.parse_page,meta={'Url':url,'Company': company,'Booth_Number': booth_num) 
 
def parse_page(self,response): 
    company = response.meta.get('Company') 
    booth_num = response.meta.get('Booth Number') 
    website = response.xpath('//a[@class="aa-BoothContactUrl"]/text()').extract_first() 
 
    yield {'Company': company,'Booth Number': booth_num, 'Website': website} 

编辑: 注释行 allowed_domains 让爬虫也可以在其他域上工作。

在 https://stackoverflow.com/a/52792350 回复您的代码