Bash - Extract region from AWS arn

Question

Very simple task: extract the region from an AWS arn.

Example:

arn:aws:lambda:eu-west-2:12345678912:layer:my-awsome-layer:3

I need to extract eu-west-2

I have a working regex for this: ^(?:[^:]+:){3}([^:]+).*

I tried this command, but it returns the entire string:

echo "arn:aws:lambda:eu-west-2:12345678912:layer:my-awsome-layer:3" | grep -oP '^(?:[^:]+:){3}([^:]+).*'

output: arn:aws:lambda:eu-west-2:12345678912:layer:my-awsome-layer:3

What is wrong with the above?

Why not just a `cut -d':' -f4`? Why do you need regex at all? — Inian, Nov 23 '22 at 15:04
@Thefourthbird I sort of realised that by looking at the output on regex101, but I still would not know how to get that out :D — Kappacake, Nov 23 '22 at 15:59
There is an example here https://stackoverflow.com/questions/1891797/capturing-groups-from-a-grep-regex — The fourth bird, Nov 23 '22 at 16:02
You can refer : https://stackoverflow.com/questions/22727107/how-to-find-the-last-field-using-cut — buttercup, Nov 23 '22 at 17:11

score 2 · Answer 1 · answered Nov 23 '22 at 15:16

2

Thanks for @Inian for the quick and good answer in the comments:

echo "arn:aws:lambda:eu-west-2:12345678912:layer:my-awsome-layer:3" | cut -d':' -f4

Does the trick.

output: eu-west-2

answered Nov 23 '22 at 15:16

Kappacake

score 2 · Answer 2 · answered Nov 23 '22 at 16:55

2

A more flexible approach:

grep -oP '\w{2}-\w+-\d+'

answered Nov 23 '22 at 16:55

Gilles Quénot

score 2 · Answer 3 · answered Nov 23 '22 at 21:27

You get back the whole line because you are using .* at the end of the pattern.

As you are using grep -oP you can make use of \K to forget what is matched so far:

grep -oP '^(?:[^:]+:){3}\K[^:]+'

An alternative using awk setting : as the field separator and printing the fourth field:

awk -F: '{print $4}'

Or using sed and replace with capture group 2, as capture group 1 is repeated 3 times:

sed 's/^\([^:]\+:\)\{3\}\([^:]\+\).*/\2/'

The examples will output:

eu-west-2

3 Answers3