[SOLVED] Linux bash script to extract IP address

Linux bash script to extract IP address

I want to make big script on my Debian 7.3 ( something like translated and much more new user friendly enviroment ). I have a problem. I want to use only some of the informations that commands give me. For example my ifconfig looks like:

eth0      Link encap:Ethernet  HWaddr 08:00:27:a3:e3:b0  
          inet addr:192.168.1.103  Bcast:192.168.1.255  Mask:255.255.255.0
          inet6 addr: fe80::a00:27ff:fea3:e3b0/64 Scope:Link
          UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1
          RX packets:1904 errors:0 dropped:0 overruns:0 frame:0
          TX packets:2002 errors:0 dropped:0 overruns:0 carrier:0
          collisions:0 txqueuelen:1000 
          RX bytes:1309425 (1.2 MiB)  T

I want to display only the IP address in line: echo "Your IP address is: (IP_ADDRESS )". Is there any command that allow me to do such a thing, to search in stream for informations I want to get?. I know about grep and sed but I am not really good with them.

Edit: Firstly to say thank you for helping me with this problem, now I know much more. Secondly to say project is in progress. If anyone would be interested in it just pm me.

Solution

To just get your IP address:

echo `ifconfig eth0 2>/dev/null|awk '/inet addr:/ {print $2}'|sed 's/addr://'`

This will give you the IP address of eth0.

Edit: Due to name changes of interfaces in recent versions of Ubuntu, this doesn't work anymore. Instead, you could just use this:

hostname --all-ip-addresses or hostname -I, which does the same thing (gives you ALL IP addresses of the host).