Question

我在这方面相当天真。我不确定为什么我的连接超时。提前谢谢。

#!/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)

此代码生成的追溯

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

Answer 1

您无需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。虽然值有点低，但是将其增加到10-15秒有望成功。

首先，这样做：

s = socket.socket()

然后：

s.settimeout(10)

你应该在连接上使用“try：”，并添加：

except socket.error as socketerror:
    print("Error: ", socketerror)

这将在输出中显示系统错误消息并处理异常。

修改后的代码版本：

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)

Answer 2

将192.168.95.148更改为127.0.0.1（localhost - 连接到您自己）并在同一台计算机上运行this program。这样你就可以连接到。

Answer 3

它是一个伪造的地址，我想如果您想使其正常工作，您正在阅读暴力的python 键入socket.gethostname（）而不是伪造的IP地址，它将连接到您的计算机。这是我的代码：

    def Socket_to_me(ip,port,time):
        try:
            s = socket.socket()
            s.settimeout(time)
            s.connect((ip,port))
            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) 

call the method by:
Socket_to_me(socket.gethostname(),1234,10)

python socket.connect - ＆gt;超时为何？

3 个答案: