socket.error

[kataibrengsek]

dear sir,

when i am running the command : python -m SimpleHTTPServer 80 in my console
i got this message :
root@bt:~# python -m SimpleHTTPServer 80
Traceback (most recent call last):
File "/usr/lib/python2.5/runpy.py", line 95, in run_module
filename, loader, alter_sys)
File "/usr/lib/python2.5/runpy.py", line 52, in _run_module_code
mod_name, mod_fname, mod_loader)
File "/usr/lib/python2.5/runpy.py", line 32, in _run_code
exec code in run_globals
File "/usr/lib/python2.5/SimpleHTTPServer.py", line 218, in <module>
test()
File "/usr/lib/python2.5/SimpleHTTPServer.py", line 214, in test
BaseHTTPServer.test(HandlerClass, ServerClass)
File "/usr/lib/python2.5/BaseHTTPServer.py", line 570, in test
httpd = ServerClass(server_address, HandlerClass)
File "/usr/lib/python2.5/SocketServer.py", line 330, in __init__
self.server_bind()
File "/usr/lib/python2.5/BaseHTTPServer.py", line 101, in server_bind
SocketServer.TCPServer.server_bind(self)
File "/usr/lib/python2.5/SocketServer.py", line 341, in server_bind
self.socket.bind(self.server_address)
File "<string>", line 1, in bind
socket.error: (98, 'Address already in use')


anyone's ever got this problem?
hope can share.
tks

[espreto]

Your question has nothing to do with Backtrack!
More follows the link That Will help you! As a tip, always read the documentation for the software you want to use!

20.19. SimpleHTTPServer — Simple HTTP request handler &mdash; Python v2.7 documentation

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