下面的代码按预期工作,运行正常:
import socketio
import eventlet
port = 5000
sio = socketio.Server()
app = socketio.WSGIApp(sio, static_files={})
@sio.event
def connect(sid, environ):
print('Connect')
eventlet.wsgi.server(eventlet.listen(('', port)), app) # Note this line
但是,一旦最后一行被包装到函数中,就会发生错误
import socketio
import eventlet
port = 5000
sio = socketio.Server()
app = socketio.WSGIApp(sio, static_files={})
@sio.event
def connect(sid, environ):
print('Connect')
def run():
eventlet.wsgi.server(eventlet.listen('', port), app) # Note this line
run()
这是完整的错误消息:
Traceback (most recent call last):
File "/home/thatcoolcoder/coding/micro-chat/tester3.py", line 16, in <module>
run()
File "/home/thatcoolcoder/coding/micro-chat/tester3.py", line 14, in run
eventlet.wsgi.server(eventlet.listen('', port), app)
File "/usr/lib/python3.9/site-packages/eventlet/convenience.py", line 49, in listen
sock = socket.socket(family, socket.SOCK_STREAM)
File "/usr/lib/python3.9/site-packages/eventlet/greenio/base.py", line 136, in __init__
fd = _original_socket(family, *args, **kwargs)
File "/usr/lib/python3.9/socket.py", line 232, in __init__
_socket.socket.__init__(self, family, type, proto, fileno)
OSError: [Errno 97] Address family not supported by protocol
我怎样才能防止这种情况?我正在构建一个小型聊天应用程序,为了保持整洁,我需要在一个函数(特别是一个类方法)中创建并运行服务器
问题在这一行:
此行应为:
更新代码:
输出:
说明:
eventlet.listen()
是eventlet.wsgi.server()
的param
,而eventlet.listen()
表示监听哪个地址和端口('', 8000)
组合了地址和端口。如果我们不设置第一个参数,它将默认为0.0.0.0如果我们设置
localhost
,它将是回溯地址127.0.0.1
,我们还可以设置计算机的IP地址相关问题 更多 >
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