I am trying to use Bash on CentOS 6.4 to retrieve the network interface name attached to an IP address using AWK. I have a bit of command from a Solaris box, but I'm not sure how to convert it to Linux output.
The command looks like this:
ifconfig -a | awk '
$1 ~ /:/ {split($1,nic,":");
lastif=sprintf("%s:%s",nic[1],nic[2]);}
$2 == "'$1'" { print lastif ; exit; }
'
Its part of a script, so it takes commandline argument like monitor.sh x.x.x.x y.y.y.y and it uses the first x.x.x.x to get the interface name, then makes $1 == $2 so then it can ping y.y.y.y later. I'm guessing that in Solaris the ifconfig -a output is different than CentOS. I can get the interface name if the IP and interface are on the same line, but in linux, they're on two different lines. Any ideas.
I don't have CentOS, but in RHEL, IP address is listed as inet address. I believe they should be same.
The following command should give you the interface name which has a IP address.
export iface=$(ifconfig | grep -B1 "inet addr:x.x.x.x" | awk '$1!="inet" && $1!="--" {print $1}')
echo "$iface" # To get the interface name for x.x.x.x ip
And this one should show the IP including localhost :
ifconfig | grep "inet addr:" | sed -e 's/addr:/addr: /g' | awk '{print $3}'
geting ifname for 127.0.0.1 (or any other IP)
ifconfig | awk '/127.0.0.1/ {print $1}' RS="\n\n"
lo
getting ip:
ifconfig | awk -F"[ :]+" '/inet addr:/ {print $4}'
Post the output of ifconfig, and I can help you fine tune for your OS
Related
ipconfig does not exist anymore as a command available in Ubuntu 20.04 and later I assume. The new command is just ip. When I run the ip address command I get the entire list of all devices and ip addresses associated. I want just the eth0 device and public ip 4 address associated.
I want just the bare ip address octets only. I want this to work on both Linux and Mac OS.
I found this pipe of cut and sed to work fine to get what I want:
on linux:
ip a | grep eth0 | cut -d " " --fields=6 | sed '2q;d' | awk -F'/' '{print $1}'
on BSD / Darwin / Mac OS:
ip a | grep en0 | tr -s 'inet' ' ' | sed '2q;d' | tr -s '' | awk -F' *? *' '{print $2}' | awk -F '/' '{print $1}'
which results in just the bare ip address I needed. I had to do some trial and error on what field column I actual needed. This probably could be more generalized, but this works for my use case.
Added a public git to just curl and run from anywhere like:
curl -L https://cutt.ly/UUYcT1r | /bin/bash
I want to read IP address of all interface and set it to no_proxy variable in centos machine.
i can do it manual by running ifconfig
this is the ip address in one of my vagrant box,
192.168.10.2
10.0.1.13
192.168.84.18
but i have around 13 boxes and ips are dynamically set everytime box is brought up.
i tried,
ifconfig | grep 192* and it gives me ip but not of all the interfaces available.
how can i set all the interface ip and assign them to no_proxy variable?
you can use awk with grep to get the ip address in your CentOS machines, and then tr command to remove \n
noip="$(ifconfig | grep inet | awk '{print $2","}' | tr -d '\n')"
it will give you,
192.168.10.2,10.0.1.13,192.168.84.18,
export it as follows including localhost and loopback address, i.e. 127.0.0.1,
export no_proxy=${noip}localhost,127.0.0.1
So your complete code will be,
noip="$(ifconfig | grep inet | awk '{print $2","}' | tr -d '\n')"
export no_proxy=${noip}localhost,127.0.0.1
I grabbed regular expression from https://www.brianparsons.net/FindIPAddresseswithawk/
~$ ips=$(ifconfig | awk '{match($0,/[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+/); if(RLENGTH > 0) { ip = substr($0,RSTART,RLENGTH); print ip}}')
~$ echo $ips
127.0.0.1 10.65.240.107
I have to store the return of this command into an array
I tried this :
my_array=( $(/sbin/ifconfig | grep 'inet addr:' | cut -d: -f2 | awk '{ print $1}' | grep -v '127.0.0.1'))
but it gives me the error below:
/bin: is a directory
You could remove parenthesis as below. You can also use ip command, its newer version and more powerful than ifconfig.
my_array=$(ip addr show | grep -Po 'inet \K[\d.]+' | grep -v 127.0.0.1)
In my GNU bash 4.4.0(1), ifconfig is not giving me inet addr: but just inet like this:
inet 10.0.2.15 netmask 255.255.255.0 broadcast 10.0.2.255
inet 127.0.0.1 netmask 255.0.0.0
In any case the error you receive doesn't make sense to me.
I can get the IP address combining grep & awk like this:
sbin/ifconfig | grep -v "127.0.0.1" |grep 'inet ' |awk -F" " '{print $2}'
(awk -F" " : Field Seperator. Field Seperator in my case is space.)
Also this works for me , due to the fact that 127.0.0.1 goes to the end:
/sbin/ifconfig |grep -m1 'inet '| awk -F" " '{print $2}
(grep -m1 = stop after first match)
PS: If you have plans to append your array , you need to use my_array+=(...)
I'm wondering how I can query by IP address using sed, and it will show which interface name that is using it.
For example..
ipconfig -a | grep 10.0.0.10
I would expect it to come back with ETH0
ifconfig | grep -B1 10.0.0.10 | grep -o "^\w*"
You should use this comand :
ifconfig | grep -B1 "inet addr:10.0.0.10" | awk '$1!="inet" && $1!="--" {print $1}'
Hope this help !
ip -br -4 a sh | grep 10.0.0.10 | awk '{print $1}'
If you want sed specific solution you may try this. Its little hard to digest how it works , but finally this combination works.
ifconfig | sed -n '/addr:10.0.0.10/{g;H;p};H;x' | awk '{print $1}'
If you want to take it as an argument via script use "$1" or so instead if 10.0.0.10.
Sed manual for reference : http://www.gnu.org/software/sed/manual/sed.html#tail
I have a bash script that runs on a variety of different Ubuntu Linux machines. Its job is to find out the LAN IPv4 address of the localhost.
The script is using
ip addr show eth0 | sed -n '/inet /{s/^.*inet \([0-9.]\+\).*$/\1/;p}'
which is fine, but some machines for some reason use eth1 instead of eth0. I would like to be able to discover the LAN iface name, so I can substitute it in here instead of eth0.
Of course, if you can come up with a different oneliner that does the same thing, all good.
The main NIC will usually have a default route. So:
ip -o -4 route show to default
The NIC:
ip -o -4 route show to default | awk '{print $5}'
The gateway:
ip -o -4 route show to default | awk '{print $3}'
Unlike ifconfig, ip has a consistent & parsable output. It only works on Linux; it won't work on other Unixen.
Not sure if this helps, but it seems that ip route get will show which interface it uses to connect to a remote host.
ubuntu#ip-10-40-24-21:/nail/srv/elasticsearch$ ip route get 8.8.8.8
8.8.8.8 via <gateway address> dev eth0 src <eth0 IP Address>
of course you could automate that in shell script with something like,
ip route get 8.8.8.8 | awk '{ print $NF; exit }'
Most recenttly systemd/udev has automatically started to assign interface names for all local Ethernet, WLAN and WWAN interfaces to something that we're all accustomed to . This is a departure from the traditional interface naming scheme ("eth0", "eth1", "wlan0", ...) .. now we have to check first what the local interface name is before we can use it while previously we it was a pretty accurate guess that "eth0" was the right name. What you're asking for is the network NAME .. Here's a small script to solve the problem
Use "ip route get 8.8.8.8 " to figure out which ACTIVE interface has the route to internet ( or currently being used )
Output should look like :
8.8.4.4 via 10.10.1.1 dev enp0s3 src 10.10.1.118
cache
Use awk to print the 5th text block for Interface NAME
]# ip route get 8.8.8.8 | awk -- '{print $5}'
Output : enp0s3
Use awk to print the 7th text block for Interface Address
]# ip route get 8.8.8.8 | awk -- '{print $7}'
Output : 10.10.1.118
How about searching for the string inet and brd (for broadcast)? That would give you:
ip addr show|egrep '^ *inet'|grep brd|awk -- '{ print $2; }'|sed -e 's:/[0-9]*$::'
Note that I'm using more commands than necessary; you can probably achieve the same thing with sed and a more complex regexp but I prefer a command that makes it obvious by which steps I arrive at the result.
If you want to run it in a single command, I suggest to try awk:
ip addr show|awk -- '$1 == "inet" && $3 == "brd" { split($2,a,"/"); print a[1]; }'
which isn't much longer than the sed version but more readable.
+1 Slightly more readable:
ip addr show | awk '$1 == "inet" && $3 == "brd" { sub (/\/.*/,""); print $2 }'
Believe it or not, there is no standard, easy way to get this information. There is no standard give me the current IP and Interface Name command. There isn't even a standard format for the information returned by ifconfig.
I was going to recommend forgoing pure shell and go with a scripting language like Python where you can do this:
import socket
socket.gethostbyname(socket.gethostname())
Except it doesn't work on most Linux systems because there's usually an entry in the /etc/host file pointing to 127.0.0.1, the loopback address. Perl has the same issues.
You have a firm grasp on the scripting involved, and you've seen the issues. The only thing I can recommend is to test this on each machine you're going to run it on, and see what pops out. There isn't going to be a general purpose one liner that works with all operating systems, or even with different systems on the same operating system because of the way the network is setup and the way interfaces may be named by each location.
I'd still like to know if there was an easier way to do this, but this is my workaround: I know the LAN subnet, so...
ip addr show | grep "inet 10.67.5." \
| sed -n '/inet /{s/^.*inet \([0-9.]\+\).*$/\1/;p}'
1) This one print only interface names (I needed that for handing upcoming and downcoming PPP links):
for i in $( ifconfig | grep 'ppp' | awk '{print $1}' );
do
printf "$i "; ## Or echo
done
Result:
ppp0 ppp1 ppp2
2) This one prints interface names and IP
declare -a IPADDR
index=0
for i in $( ifconfig | grep 'inet addr' | awk '{print $2}'| sed 's#addr:##g' );
do
IPADDR[$index]=$i
let "index += 1"
done
index=0
for i in $( ifconfig | grep 'ppp' | awk '{print $1}' );
do
echo $i
let "index += 1"
done
Result:
ppp0 addr:IP
ppp1 addr:IP
ppp2 addr:IP