Linux bash script to extract IP address

Question

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 informa

Answer 1

Just a note, since I just spent some time trouble-shooting a botched upgrade on a server. Turned out, that (years ago) I had implemented a test to see if dynamically added interfaces (e.g. eth0:1) were present, and if so, I would bind certain proggis to the 'main' IP on eth0. Basically it was a variation on the 'ifconfig|grep...|sed... ' solution (plus checking for 'eth0:' presence).
The upgrade brought new net-tools, and with it the output has changed slightly:

old ifconfig:

eth0      Link encap:Ethernet  HWaddr 42:01:0A:F0:B0:1D
          inet addr:10.240.176.29  Bcast:10.240.176.29  Mask:255.255.255.255
          UP BROADCAST RUNNING MULTICAST  MTU:1460  Metric:1
          ...<SNIP>

whereas the new version will display this:

eth0: flags=4163<UP,BROADCAST,RUNNING,MULTICAST>  mtu 1460
      inet 10.240.212.165  netmask 255.255.255.255  broadcast 10.240.212.165
      ...<SNIP>

rendering the hunt for 'eth0:' as well as 'inet addr:' search busted (never mind interfaces called 'em0','br0' or 'wlan0'...). Sure you could check for 'inet ' (or 'inet6'), and make the addr: part optional, but looking closer, you'll see that more or less everything has changed, 'Mask' is now 'netmask',...

The 'ip route ...' suggestion's pretty nifty - so maybe:

_MyIP="$( ip route get 8.8.8.8 | awk 'NR==1 {print $NF}' )"
if [ "A$_MyIP" == "A" ]
then
    _MyIPs="$( hostname -I )"
    for _MyIP in "$_MyIPs"
    do
        echo "Found IP: \"$_MyIP\""
    done
else
    echo "Found IP: $_MyIP"
fi

Well, something of that sort anyway. Since all proposed solutions seem to have circumstances where they fail, check for possible edge cases - no eth, multiple eth's & lo's, when would 'hostname -i' fail,... and then decide on best solution, check it worked, otherwise 2nd best.

Cheers 'n' beers!

Answer 2

In my opinion the simplest and most elegant way to achieve what you need is this:

ip route get 8.8.8.8 | tr -s ' ' | cut -d' ' -f7

---

ip route get [host] - gives you the gateway used to reach a remote host e.g.:

8.8.8.8 via 192.168.0.1 dev enp0s3  src 192.168.0.109

---

tr -s ' ' - removes any extra spaces, now you have uniformity e.g.:

8.8.8.8 via 192.168.0.1 dev enp0s3 src 192.168.0.109

---

cut -d' ' -f7 - truncates the string into ' 'space separated fields, then selects the field #7 from it e.g.:

192.168.0.109

Answer 3

Take your pick:

$ cat file
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

$ awk 'sub(/inet addr:/,""){print $1}' file
192.168.1.103

$ awk -F'[ :]+' '/inet addr/{print $4}' file
192.168.1.103

Answer 4

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).

Answer 5

If You want to use only sed to extract IP address:

ifconfig eth0 2>/dev/null sed -n 's/.*[[:space:]]\([[:digit:]][[:digit:]]*\.[[:digit:]][[:digit:]]*\.[[:digit:]][[:digit:]]*\.[[:digit:]][[:digit:]]*\).*/\1/p'

Answer 6

  /sbin/ifconfig eth0 | grep 'inet addr:' | cut -d: -f2 | awk '{ print $1}'