[luke14free]

Hi, I have tried many and many times to adapt the pc socket.py and similar modules to pys60 but i failed all the times...I need to create a server (better if ftp) on my phone. I'd have 3 question for you:

1)How to get current device ip for the server?(socket.py doesn't seem to work)
2)How to create an ftp/http server on my phone?
3)If isn't possible yet to create ftp/http servers on phones, do you think that there is another possibility to send data from device to device (not sms of course ) or from pc to device-device to pc by web/wap?

Thanks a lot

[asteroidm]

dude..,

i think "racoon" webserver is the answer to all your queries.. []

[luke14free]

Hmm...I am trying to use a socket module to create a self server...

import socket
S = socket.socket ( socket.AF_INET, socket.SOCK_STREAM )
S.bind ( ( 'localhost', 80 ) )
S.listen ( 1 )
while True:
   channel, details = S.accept()
   print 'Connected with', details
   print channel.recv ( 100 )
   channel.send ( 'Server online' )
   channel.close()

This code should be good because it works on my pc but on the phone i get this error:

Traceback (most recent call last):
  File "<console>", line 1 in ?
  File "<string>", line 1 in bind
error: (22, 'Invalid argument')

Why??
Is it a bug? Where could I find s60 socket docs?

Kind Regards,
Luke14free

[lfd]

Hi luke14free,

Socket doc is here http://www.python.org/doc/2.2.2/lib/lib.html

Check raccon, this is a Apache webserver for Symbian. You might be interested, it depends what you want to do.

LFD

[lfd]

Hi luke14free,

Little erratum: check also section 8.2 (S60 extensions to standard socket module) of the PyS60 documentation http://switch.dl.sourceforge.net/sou...doc_1_3_21.pdf

Cheers,

LFD

[janimr]

luke14free,

S.bind ( ( 'localhost', 80 ) )

or '127.0.0.1' won't work. You have to use correct interface IP address instead of localhost.

- janimr

[jplauril]

Originally Posted by luke14free

Hi, I have tried many and many times to adapt the pc socket.py and similar modules to pys60 but i failed all the times...I need to create a server (better if ftp) on my phone. I'd have 3 question for you:

1)How to get current device ip for the server?(socket.py doesn't seem to work)
2)How to create an ftp/http server on my phone?
3)If isn't possible yet to create ftp/http servers on phones, do you think that there is another possibility to send data from device to device (not sms of course ) or from pc to device-device to pc by web/wap?

Thanks a lot

The new 1.3.22 release will include a way to obtain the IP address associated with a given access point. However, this will not do you any good with cellular access points, in practice you can only use this over WLAN, since most operators block incoming TCP. If you know of exceptions then please let us know.

-Jukka

[.iqqmuT]

I want to create a TCP/IP server into my phone using WLAN in my private network. Is there any way to find out the current ip?

This is from python documentations:

If you want to know the current machine's IP address, you may want to use gethostbyname(gethostname()).

http://docs.python.org/lib/module-socket.html

My phone returns string with asterisks, like '****************'.
Well, this wasn't surprise because in Release Notes of PyS60 it states:

Known issues ...

1446351 227: socket.gethostbyname('localhost') inconsistencies

- This call returns inconsistent results on different S60 platforms
and depending on the state, and should not be relied on.

http://sourceforge.net/project/shown...roup_id=154155

Is there any other way to find out my IP?

This works if my IP is for example 172.16.0.23:

import socket
S = socket.socket ( socket.AF_INET, socket.SOCK_STREAM )
S.bind ( ( '172.16.0.23', 2000 ) )
S.listen ( 1 )
channel, details = S.accept()
print 'Connected with', details
print channel.recv ( 100 )
channel.send ( 'Server online' )
channel.close()

Of course I could create a simple TCP/IP service which returns back the IP of the client and use it from the phone before starting the server but it requires having this server installed into the network.