To accepts socket connetction you need always listen some port. Othewise
you will be trying hit a bulls eye when sending data to this port (if you
would listen port only when view was called) View is a function it's not a
webserver that runs and listen to port.
You need to decouple working with sockets from view and place it somewere
outside django
2011/12/27 Vovk Donets <donets.vladi...@gmail.com>
> For what you need this?
> Maybe you should consider using Celery for doing background jobs.
>
>
> 2011/12/23 Kay <fcapba1...@gmail.com>
>
>> if I have easy asyncore socket code as follow:
>>
>>
>> --------------------------------------------------------------------------------------------------------------
>>
>> import socket
>> import asyncore
>>
>> class asysocket(asyncore.dispatcher):
>>
>> def __init__(self,host,port):
>> asyncore.dispatcher.__init__(self)
>> self.create_socket(socket.AF_INET, socket.SOCK_STREAM)
>> self.set_reuse_addr()
>> self.bind((host, port))
>> self.listen(10)
>> print 'server is waiting for socket connection...'
>>
>> def handle_accept(self):
>> new,addr = self.accept()
>> print 'received from address', addr
>> data = new.recv(1024)
>> print repr(data)
>>
>> def handle_close(self):
>> self.close()
>>
>> server = asysocket('127.0.0.1',1234)
>> asyncore.loop()
>>
>>
>> ----------------------------------------------------------------------------------------------------------------
>>
>> then how to use this in my django project views.py to let it
>> automatically work for receiving connection
>>
>> while i start the project as "python manage.py runserver [port]"??
>>
>> thanks
>>
>>
> --
> *Vovk Donets*
> python developer
>
> skype: suunbeeam
> icq: 232490857
> mail: donets.vladi...@gmail.com
> www: jetfix.ru
>
>
--
*Vovk Donets*
python developer
skype: suunbeeam
icq: 232490857
mail: donets.vladi...@gmail.com
www: jetfix.ru
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