6

伙计们，我是套接字编程的新手以下程序是一个客户端程序，它从服务器请求文件，但我收到如下所示的错误.. 我的输入是 GET index.html，代码是任何人都可以解决这个错误...？

#!/usr/bin/env python

import httplib
import sys


http_server = sys.argv[0]

conn = httplib.HTTPConnection(http_server)

while 1:
cmd = raw_input('input command (ex. GET index.html): ')
cmd = cmd.split()

if cmd[0] == 'exit': 
    break


conn.request(cmd[0],cmd[1])


rsp = conn.getresponse()


print(rsp.status, rsp.reason)
data_received = rsp.read()
print(data_received)

conn.close()





input command (ex. GET index.html): GET index.html
Traceback (most recent call last):
File "./client1.py", line 19, in <module>
conn.request(cmd[0],cmd[1])
File "/usr/lib/python2.6/httplib.py", line 910, in request
self._send_request(method, url, body, headers)
File "/usr/lib/python2.6/httplib.py", line 947, in _send_request
self.endheaders()
File "/usr/lib/python2.6/httplib.py", line 904, in endheaders
self._send_output()
File "/usr/lib/python2.6/httplib.py", line 776, in _send_output
self.send(msg)
File "/usr/lib/python2.6/httplib.py", line 735, in send
  self.connect()
 File "/usr/lib/python2.6/httplib.py", line 716, in connect
  self.timeout)
File "/usr/lib/python2.6/socket.py", line 500, in create_connection
for res in getaddrinfo(host, port, 0, SOCK_STREAM):
socket.gaierror: [Errno -2] Name or service not known
4

2 回答 2

3

sys.argv[0] 不是你想象的那样。sys.argv[0] 是程序或脚本的名称。脚本的第一个参数是 sys.argv[1]。

于 2013-09-19T14:50:07.960 回答
2

问题是第一项sys.argv是脚本名称。所以你的脚本实际上是使用你的文件名作为主机名。将第 5 行更改为：

http_server = sys.argv[1]

更多信息在这里。

于 2013-09-19T14:48:42.297 回答