Parse command arp -a for only ip

Question

I need to parse the command arp -a to get only the ip of the device. Right now I have

arp -a | awk '{print $2}' | head -1

However this gives me

(192.168.1.71)

and I must remove the ( ) from the output. How can this be done?

welcome to U&L, what do you meant by **get the only ip of the device** ? on most unix/linux system `arp -a` will give a random list of recently used IPs and MACs. — Archemar, Jul 25 '19 at 06:36

score 2 · Accepted Answer · answered Jul 22 '19 at 16:06

2

You could use the characters () as field separators instead of whitespace:

arp -a | awk -F'[()]' '{print $2}' | head -1

answered Jul 22 '19 at 16:06

user4556274

HarshaD · Answer 2 · 2019-07-24T13:23:13.030

1

using cut :

arp -a | awk '{print $2}' | head -1 | cut -d "(" -f 2 | cut -d ")" -f 1

Updating the answer as suggested in the comment:

cut explanation:

-d - acts as delimiter and splits the string on first occurrence of (

-f - selects the specified field. Here selects the 2nd part which would be x.x.x.x)

We give this output as input to next cut using |

cut -d ")" -f 1 - This again splits the input on ) and we select the first part i.e. Just the IP address.

edited Jul 24 '19 at 13:23

answered Jul 22 '19 at 16:00

HarshaD

It's considered a good practise to explain a command so future users can learn something and adapt it to their needs. – Philippos Jul 24 '19 at 10:18
@Philippos - Sure will keep in mind. Updated the answer. Thanks! – HarshaD Jul 24 '19 at 13:23

score 0 · Answer 3 · answered Jul 22 '19 at 22:43

0

tr is good at things like deleting unwanted characters:

$ tr -d '()' <<< '(192.168.1.1)'
192.168.1.1

So all told, it would be:

arp -a | awk '{print $2}' | head -1 | tr -d '()'

answered Jul 22 '19 at 22:43

Jim L.

3 Answers3