如何优雅退出-套接字错误处理

Question

嗨，我正在写一个用户输入url的脚本。我应该验证url，并让用户知道如果url是valid.If我故意键入错误的url我得到了一个错误的消息gaierror，但我已经把错误处理代码在这里，使我的程序得到一条消息“无法连接”喜欢的代码行“如果exit.Seems是无：-print‘无法打开套接字’不是在下面的程序中执行。我是在套接字编程新手。我想我正在做的事情wrong.Could任何人请帮助我提前here.Thanks！

#!/usr/bin/env python
import socket
import sys


def hello(c):
  t=c
  port = 80
  slash=t.split("/")
  basename="/".join(slash[2:3])
  pathname="/"+"/".join(slash[3:])
  sock=None

  for res in socket.getaddrinfo(basename, port, socket.AF_UNSPEC, socket.SOCK_STREAM):
     af, socktype, proto, canonname, sa = res
     try:
        sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
     except socket.error:
        sock= None
        continue
     try:
        sock.connect((basename, port))
     except socket.error:            
        sock= None
        continue

  if sock is None:
    print 'could not open socket'###I dont get this printed 
    sys.exit(0)   

  sock.send("HEAD %s HTTP/1.0\r\n\r\n" % pathname)
  l=sock.recv(12)
  return l
  sys.exit(1)
  sock.close()


``I call the hello function from here``
k= hello("https://mail.yah")`purposefully wrong to verify error handle
print k
newsplit=k.split()
print newsplit
insplit=eval(newsplit[1])
print insplit
if insplit==404:
 print "bad"
else:
 print "ok"

我得到以下错误消息：

 Traceback (most recent call last):
 File "C:/myserver/cgi-bin/domain.py", line 40, in <module>
 k= hello("https://mail.yah")
 "C:/myserver/cgi-bin/domain.py", line 13, in hello
 for res in socket.getaddrinfo(basename, port, socket.AF_UNSPEC, socket.SOCK_STREAM):
 gaierror: [Errno 11004] getaddrinfo failed

Answer 1

您可能希望将该socket.getaddrinfo(basename, port, socket.AF_UNSPEC, socket.SOCK_STREAM)包装在try / except socket.gaierror块中，这应该可以做到这一点：

try:
    for res in socket.getaddrinfo(basename, port, socket.AF_UNSPEC, socket.SOCK_STREAM):
         af, socktype, proto, canonname, sa = res
         # stuff
except socket.gaierror:
    print 'getaddrinfo failed!'

实际上，是socket.getaddrinfo(basename, port, socket.AF_UNSPEC, socket.SOCK_STREAM)导致了这个错误！