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[Michel Leunen <618868@xxxxxxxxxxxxxxxxxx>]

Yes, there is another way: using the command ip from the net-tools package instead of ifconfig. This package is installed by default in Ubuntu. Don't know for other distributions.
Use it like this:

$ ip addr show

On my machine, it prints this:

$ ip addr show
1: lo: <LOOPBACK,UP,LOWER_UP> mtu 16436 qdisc noqueue state UNKNOWN 
    link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00
    inet 127.0.0.1/8 scope host lo
    inet6 ::1/128 scope host 
       valid_lft forever preferred_lft forever
2: eth0: <NO-CARRIER,BROADCAST,MULTICAST,UP> mtu 1500 qdisc pfifo_fast state DOWN qlen 1000
    link/ether 00:25:b3:c0:b6:86 brd ff:ff:ff:ff:ff:ff
3: wlan0: <BROADCAST,MULTICAST,UP,LOWER_UP> mtu 1500 qdisc mq state UP qlen 1000
    link/ether 00:21:6a:6c:de:18 brd ff:ff:ff:ff:ff:ff
    inet 192.168.1.8/24 brd 192.168.1.255 scope global wlan0
    inet6 fe80::221:6aff:fe6c:de18/64 scope link 
       valid_lft forever preferred_lft forever
4: vboxnet0: <BROADCAST,MULTICAST> mtu 1500 qdisc noop state DOWN qlen 1000
    link/ether 0a:00:27:00:00:00 brd ff:ff:ff:ff:ff:ff

You can narrow it to what is interresting with:

$ ip -o -4 addr show
1: lo    inet 127.0.0.1/8 scope host lo
3: wlan0    inet 192.168.1.8/24 brd 192.168.1.255 scope global wlan0

So basically your function stays the same.

I've tried to use the python socket module directly but it gives an ip
address even if the connection is lost for some reason:

>>> import socket
>>> socket.gethostbyname_ex(socket.gethostname())[2][0]
'192.168.1.8'

So I suppose your best bet is to go the 'ip addr show' way.

-- 
sysmonitor screenlet fails to find the computer's IP address
https://bugs.launchpad.net/bugs/618868
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