小编Col*_*rld的帖子

我试图通过关键字刮取这个xml页面的链接,但urllib2给我带来的错误,我无法解决python3 ...

from bs4 import BeautifulSoup
import requests
import smtplib
import urllib2
from lxml import etree
url = 'https://store.fabspy.com/sitemap_products_1.xml?from=5619742598&to=9172987078'
hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
       'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
       'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
       'Accept-Encoding': 'none',
       'Accept-Language': 'en-US,en;q=0.8',
       'Connection': 'keep-alive'}
proxies = {'https': '209.212.253.44'}
req = urllib2.Request(url, headers=hdr, proxies=proxies)
try:
    page = urllib2.urlopen(req)
except urllib2.HTTPError as e:
    print(e.fp.read())
content = page.read()
def parse(self, response):
    try:
        print(response.status)
        print('???????????????????????????????????')
        if response.status == 200:
            self.driver.implicitly_wait(5)
            self.driver.get(response.url)
            print(response.url)
            print('!!!!!!!!!!!!!!!!!!!!')

            # DO STUFF …

无论如何,我可以让python点击一个链接,如bit.ly链接,然后刮取结果链接？当我抓取某个页面时,我可以抓取的唯一链接是重定向的链接,重定向到的链接是我需要的信息所在的位置.