Question

我试图通过关键字抓取这个xml页面的链接，但是urllib2给我带来了我无法修复python3的错误......

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
    except httplib.BadStatusLine:
        pass
while True:
    soup = BeautifulSoup(a.context, 'lxml')
    links = soup.find_all('loc')
    for link in links:
        if 'notonesite' and 'winter' in link.text:
            print(link.text)
            jake = link.text

我只是尝试通过代理发送urllib请求，以查看链接是否在站点地图上...

Answer 1

urllib2在Python3中不可用。您应该使用urllib.error和urllib.request：

import urllib.request
import urllib.error
...
req = (url, headers=hdr) # doesn't take a proxies argument though...
...
try:
    page = urllib.request.urlopen(req)
except urllib.error.HTTPError as e:
...

......等等。但请注意，urllib.request.Request()不会使用proxies参数。有关代理处理，请参阅the documentation。

SyntaxError：语法无效：除了urllib2.HTTPError，e：

1 个答案: