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Я хочу получить статус Url для сайтов с кодом ниже. Для одного веб-сайта (webscraper.io) у меня возникла ошибка. Мой сценарий:httplib error on socket.gaierror:
import httplib
url = "http://webscraper.io/"
if 'http' in url:
url = url.replace('http://', '').strip()
conn = httplib.HTTPConnection(url)
conn.request("GET",'')
r1 = conn.getresponse()
print 'r1.Status code=', r1.status
я получил следующие ошибки:
Traceback (most recent call last):
File "TestSatusline.py", line 23, in <module>
conn.request("GET",'')
File "/usr/lib/python2.7/httplib.py", line 1017, in request
self._send_request(method, url, body, headers)
File "/usr/lib/python2.7/httplib.py", line 1051, in _send_request
self.endheaders(body)
File "/usr/lib/python2.7/httplib.py", line 1013, in endheaders
self._send_output(message_body)
File "/usr/lib/python2.7/httplib.py", line 864, in _send_output
self.send(msg)
File "/usr/lib/python2.7/httplib.py", line 826, in send
self.connect()
File "/usr/lib/python2.7/httplib.py", line 807, in connect
self.timeout, self.source_address)
File "/usr/lib/python2.7/socket.py", line 553, in create_connection
for res in getaddrinfo(host, port, 0, SOCK_STREAM):
socket.gaierror: [Errno -2] Name or service not known
ли кто-нибудь имеет какие-либо идеи?
благодаря