Socketuri in Python pe Windows 7

#1
old_jhon

Posted 17 April 2014 - 08:19

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Salutare. Incerc sa ma conectez la un server dintr-un script Python pe Windows si primesc eroarea de mai jos. Pe Linux merge OK codul de mai jos.

Codul arata cam asa:

s=socket.socket(socket.AF_INET, socket.SOCK_RAW, socket.IPPROTO_TCP)
ip_address = '192.168.0.2'
s.connect((ip_address, 1))

Si eroarea:

s.connect((ip_address, 1))
  File "<string>", line 1, in connect
error: [Errno 10022] An invalid argument was supplied

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#2
Bucifon

Posted 17 April 2014 - 08:30

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incearca asa

s=socket.socket(socket.AF_INET, socket.SOCK_RAW, socket.IPPROTO_TCP)
ip_address = '192.168.0.2'
port = 1
s.connect((ip_address, port))

cred ca are legatura cu modul in care declari adresa IP

"The simplest way to create addresses is to use the ipaddress.ip_address() factory function, which automatically determines whether to create an IPv4... address based on the passed in value:

https://docs.python..../ipaddress.html

Edited by Bucifon, 17 April 2014 - 08:46.

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#3
old_jhon

Posted 17 April 2014 - 13:09

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Am ramas pe linux, nu are rost sa ma chinui cu Windowsul. Modulul acela ipaddress cred ca este pentru Python 3.4, eu folosesc 2.6 pe Windows. Merci pentru ajutor oricum.

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#4
dani.user

Posted 17 April 2014 - 13:12

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Posts: 30,254
Înscris: 24.02.2007

TCP e SOCK_STREAM, nu RAW.
Windows nu te lasa sa umblii cu socketuri RAW.

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Socketuri in Python pe Windows 7

#1 old_jhon Posted 17 April 2014 - 08:19

#2 Bucifon Posted 17 April 2014 - 08:30

#3 old_jhon Posted 17 April 2014 - 13:09

#4 dani.user Posted 17 April 2014 - 13:12

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#1
old_jhon

Posted 17 April 2014 - 08:19

#2
Bucifon

Posted 17 April 2014 - 08:30

#3
old_jhon

Posted 17 April 2014 - 13:09

#4
dani.user

Posted 17 April 2014 - 13:12