【发布时间】:2019-07-16 12:32:59
【问题描述】:
我有这个网址 www.example.com。我需要在scrapy Request方法中发送这个url,我怎样才能做到这一点? url_final = https://www.example.com/sch/i.html?_from=R40&_trksid=m570.l1313&_nkw=MYSEARCHKEY&_sacat=0
url = 'httpss://www.example.com/'
MYSEARCHKEY = 'MYSEARCHKEY'
yield Request(url_final, callback=self.parse_new)
【问题讨论】:
标签: python callback request scrapy-spider