返回列表的差异

Question

我似乎打赌没有得到返回值的区别列表旧网址和新网址。这个脚本应该不断地轮询，直到一个差异附加这个差异并返回这个差异。然后返回到def newresponse（）

from bs4 import BeautifulSoup
import requests
import time
old_urls = []
new_urls = []


def main():
  s = requests.session()
  url = s.get('https://www.sivasdescalzo.com/sitemaps/en/sitemap-1.xml')
  soup = BeautifulSoup(url.content, "html.parser")
  all_urls = soup.find_all("url")
  for url in all_urls:
     old_urls.append(url.find('loc').get_text())
def newresponse():
  s = requests.session()
  url = s.get('https://www.sivasdescalzo.com/sitemaps/en/sitemap-1.xml')
  soup2 = BeautifulSoup(url.content, "html.parser")
  all_newurls = soup2.find_all("url")
  for urls in all_newurls:
     new_urls.append(urls.find('loc').get_text())

def monitorchange():
  x = list(set(new_urls) - set(old_urls))
  print "looking for change"
  while True:
     s =requests.session()
     url = s.get('https://www.sivasdescalzo.com/sitemaps/en/sitemap-1.xml')
     soup3 = BeautifulSoup(url.content, "html.parser")
     if new_urls != old_urls:
         return x
         old_urls.append(x)
         continue
     elif url.status_code==403:
         print "bannned"
     else:
         time.sleep(60)

main()
newresponse()
monitorchange()

Answer 1

使用def定义函数，而不是调用它。不会调用main（）和newresponse（）函数。因此，您的旧列表和新列表将保持为空，不能显示任何差异