我有这个网址www.example.com。我需要在scrapy请求方法中发送此URL,我该如何实现? 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)
您可以使用python库进行urlencoding来创建最终URL。
from urllib.parse import urlencode
params={'_from':'R40',
'_trksid':'m570.l1313',
'_nkw':'MYSEARCHKEY',
'_sacat':0}
base_url=' https://www.example.com/sch/i.html'
final_url= '{}?{}'.format(base_url,urlencode(params))