How do I get my public ip address in a shell script?

Question

This command returns my ip address with additional information.

dig @resolver1.opendns.com myip.opendns.com
; <<>> DiG 9.6-ESV-R4-P3 <<>> @resolver1.opendns.com myip.opendns.com
; (1 server found)
;; global options: +cmd
;; Got answer:
;; ->>HEADER<<- opcode: QUERY, status: NOERROR, id: 48206
;; flags: qr rd ra; QUERY: 1, ANSWER: 1, AUTHORITY: 0, ADDITIONAL: 0

;; QUESTION SECTION: 
;myip.opendns.com.      IN  A

;; ANSWER SECTION:
myip.opendns.com.   0   IN  A   122.167.119.178

;; Query time: 199 msec
;; SERVER: 208.67.222.222#53(208.67.222.222)
;; WHEN: Fri May 18 11:46:51 2012
;; MSG SIZE  rcvd: 50

I only want to extract my ip address from this. How can I extract my ip address from the dig output?

Answer 1

Don't make this harder than it needs to be... use +short

[mpenning@Bucksnort ~]$ dig +short mike.homeunix.com
76.21.48.169
[mpenning@Bucksnort ~]$

Answer 2

If you are using bash shell this will work for you

grep -A1 "ANSWER SECTION" ip_file.txt  | awk '{if(NF==5)print $5;}'

NOTE: My assumption is you are planning to extract the ip printed after the "ANSWER SECTION"

Answer 3

dig +short myip.opendns.com @resolver1.opendns.com

From my blog: http://blog.valch.name/2016/03/17/show-your-ip/