尝试解析电子邮件的 html 时"Cannot use string pattern on bytes like object"

Question

所以我有一个脚本，我已经使用了几天，试图从我拥有的csv中获取电子邮件列表，但现在我遇到了这个障碍。这是代码：

import sys
try:
import urllib.request as urllib2
except ImportError:
import urllib2
import re
import csv
list1 = []
list2 = []
list3 = []
def addList():
with open('file.csv', 'rt') as f:
reader = csv.reader(f)
for row in reader:
for s in row:
list2.append(s)
def getAddress(url):
http = "http://"
https = "https://"
if http in url:
return url
elif https in url:
return url
else:
url = "http://" + url
return url
def parseAddress(url):
global list3
try:
website = urllib2.urlopen(getAddress(url))
html = website.read()
addys = re.findall('''[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?''', html, flags=re.IGNORECASE)
global list1
list1.append(addys)
except urllib2.HTTPError as err:
print ("Cannot retrieve URL: HTTP Error Code: "), err.code
list3.append(url)
except urllib2.URLError as err:
print ("Cannot retrive URL: ") + err.reason[1]
list3.append(url)
def execute():
global list2
addList()
totalNum = len(list2)
atNum = 1
for s in list2:
parseAddress(s)
print ("Processing ") + str(atNum) + (" out of ") + str(totalNum)
atNum = atNum + 1
print ("Completed. Emails parsed: ") + str(len(list1)) + "."

### MAIN
def main():
global list2
execute()
global list1
myFile = open("finishedFile.csv", "w+")
wr = csv.writer(myFile, quoting=csv.QUOTE_ALL)
for s in list1:
wr.writerow(s)
myFile.close
global list3
failFile = open("failedSites.csv", "w+")
write = csv.writer(failFile, quoting=csv.QUOTE_ALL)
for j in list3:
write.writerow(j)
failFile.close
main()

当我运行它时，出现此错误：

Traceback (most recent call last):
File "pagescanner.py", line 85, in <module>
main()
File "pagescanner.py", line 71, in main
execute()
File "pagescanner.py", line 60, in execute
parseAddress(s)
File "pagescanner.py", line 42, in parseAddress
addys = re.findall('''[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?''', html, flags=re.IGNORECASE)
File "/usr/lib/python3.5/re.py", line 213, in findall
return _compile(pattern, flags).findall(string)
TypeError: cannot use a string pattern on a bytes-like object

所以我发现我需要弄清楚如何将 html 字符串编码为编码的字节，而 Tyler 下面的回答帮助我做到了这一点，但现在我收到此错误：

Traceback (most recent call last):
File "/usr/lib/python3.5/urllib/request.py", line 1254, in do_open
h.request(req.get_method(), req.selector, req.data, headers)
File "/usr/lib/python3.5/http/client.py", line 1107, in request
self._send_request(method, url, body, headers)
File "/usr/lib/python3.5/http/client.py", line 1152, in _send_request
self.endheaders(body)
File "/usr/lib/python3.5/http/client.py", line 1103, in endheaders
self._send_output(message_body)
File "/usr/lib/python3.5/http/client.py", line 934, in _send_output
self.send(msg)
File "/usr/lib/python3.5/http/client.py", line 877, in send
self.connect()
File "/usr/lib/python3.5/http/client.py", line 849, in connect
(self.host,self.port), self.timeout, self.source_address)
File "/usr/lib/python3.5/socket.py", line 712, in create_connection
raise err
File "/usr/lib/python3.5/socket.py", line 703, in create_connection
sock.connect(sa)
OSError: [Errno 22] Invalid argument
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "pagescanner.py", line 39, in parseAddress
website = urllib2.urlopen(getAddress(url))
File "/usr/lib/python3.5/urllib/request.py", line 163, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python3.5/urllib/request.py", line 466, in open
response = self._open(req, data)
File "/usr/lib/python3.5/urllib/request.py", line 484, in _open
'_open', req)
File "/usr/lib/python3.5/urllib/request.py", line 444, in _call_chain
result = func(*args)
File "/usr/lib/python3.5/urllib/request.py", line 1282, in http_open
return self.do_open(http.client.HTTPConnection, req)
File "/usr/lib/python3.5/urllib/request.py", line 1256, in do_open
raise URLError(err)
urllib.error.URLError: <urlopen error [Errno 22] Invalid argument>
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "pagescanner.py", line 85, in <module>
main()
File "pagescanner.py", line 71, in main
execute()
File "pagescanner.py", line 60, in execute
parseAddress(s)
File "pagescanner.py", line 51, in parseAddress
print ("Cannot retrive URL: ") + err.reason[1]
TypeError: 'OSError' object is not subscriptable

这是否意味着列表中的某个网址不是有效的网址？我以为我终于从我的csv文件中删除了所有错误的网址，但我可能需要再看一眼

Answer 1

要回答您的问题，您只需要正确解码响应即可。 与其尝试html = website.read()html = website.read().decode('utf-8')

请参阅将字节转换为字符串

我还会推荐一些可能会让你的生活更轻松的事情。urllib.parse使处理URL变得不那么令人头疼，并且当您不可避免地在某处遇到错误时，往往会使事情更具可读性。

https://docs.python.org/3.5/library/urllib.parse.html

requests库也几乎是处理HTTP请求的黄金标准，可能有助于解决标准urllib.request编码和其他开销方面的一些混乱。

https://requests.readthedocs.io/en/master/

beautifulsoup是处理HTML的绝佳工具。

https://www.crummy.com/software/BeautifulSoup/bs4/doc/#