python socket.connect -> 为什么超时？
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python socket.connect -> timed out why?
提问by Drew Verlee
I'm fairly naive in this regard. I'm not sure why my connection is timing out. Thanks in advance.
在这方面，我相当天真。我不确定为什么我的连接超时。提前致谢。
#!/usr/bin/env python
import socket
def socket_to_me():
socket.setdefaulttimeout(2)
s = socket.socket()
s.connect(("192.168.95.148",21))
ans = s.recv(1024)
print(ans)
the trace back generated by this code
此代码生成的回溯
Traceback (most recent call last):
File "logger.py", line 12, in <module>
socket_to_me()
File "/home/drew/drewPlay/python/violent/networking.py", line 7, in socket_to_me
s.connect(("192.168.95.148",21))
File "/usr/lib/python2.7/socket.py", line 224, in meth
return getattr(self._sock,name)(*args)
timeout: timed out
采纳答案by Aleksander S
You don't need to alter the default timeouts for all new sockets, instead you can just set the timeout of that particular connection. The value is a bit low though, so increasing it to 10-15 seconds will hopefully do the trick.
您不需要更改所有新套接字的默认超时，而只需设置该特定连接的超时即可。不过该值有点低，因此将其增加到 10-15 秒有望成功。
First, do this:
首先，这样做：
s = socket.socket()
Then:
然后：
s.settimeout(10)
And you should use "try:" on the connect, and add:
您应该在连接上使用“try:”，并添加：
except socket.error as socketerror:
print("Error: ", socketerror)
This will bring up the systems error message in your output and handle the exception.
这将在您的输出中显示系统错误消息并处理异常。
Modified version of your code:
您的代码的修改版本：
def socket_to_me():
try:
s = socket.socket()
s.settimeout(2)
s.connect(("192.168.95.148",21))
ans = s.recv(1024)
print(ans)
s.shutdown(1) # By convention, but not actually necessary
s.close() # Remember to close sockets after use!
except socket.error as socketerror:
print("Error: ", socketerror)
回答by Charles Noon
Change 192.168.95.148 to 127.0.0.1 (localhost - connecting to yourself) and run this programon the same machine. That way you'll have something to connect to.
将 192.168.95.148 更改为 127.0.0.1（本地主机 - 连接到自己）并在同一台机器上运行此程序。这样你就可以连接一些东西。
