how to iterate over awk result - bash

Question

I have the following string that I want to retrieve a specific ID for eu-central-1 only:
ca-central-1:ami-aaaa,eu-central-1:ami-bbbb,eu-north-1:ami-cccc,eu-west-1:ami-dddd
so what I want as an output is: ami-bbbb
The way I am doing it right now is:
echo a-central-1:ami-aaaa,eu-central-1:ami-bbbb,eu-north-1:ami-cccc,eu-west-1:ami-dddd |
awk -F',' '{ print $2 }' |
awk -F':' '{print $2}'
The problem with this approach is that I am explicity specifying that eu-central-1 is the second ($2) result for the first awk call, but sometimes they might in different order, so I might need to iterate over this result. Is it possible to achieve this in one line, and without knowing before hand in which place in the string eu-central-1:ami-bbbb will land?

Answer 1

Use grep like so:
echo your_string | grep -Po '\beu-central-1:\K[^,]+'
Here, grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only, 1 match/line, not the entire lines.
\b : Word boundary.
\K : Pretend that the match starts at this point. Specifically, ignore the preceding part of the regex when printing the match.
[^,]+ : Any characters that are not a comma, one or more occurrences.
SEE ALSO:
grep manual

Answer 2

I'd prefer grep as in Timur Shtatland's answer. But for completeness here is an alternative:
You can set awk's record separator (linebreak by default) and then only print that record starting with eu-central-1.
awk -F: -v RS=, '$1 == "eu-central-1" { print $2 }'

Answer 3

With GNU sed or OSX/BSD sed for -E:
$ sed -E 's/(^|.*,)eu-central-1:([^,]*).*/\2/' file
ami-bbbb

Answer 4

One sed idea:
id='eu-central-1'
# desired id in middle of input string:
echo 'a-central-1:ami-aaaa,eu-central-1:ami-bbbb,eu-north-1:ami-cccc,eu-west-1:ami-dddd' | \
sed -En "s/^(.*,)*${id}:([^,]*)(,.*)*$/\2/p"
Where:
-En - enable extended regex support
^(.*,)* - [capture group #1] - matches start of line plus zero or more instances of characters ending with a comma (,)
^(.*,)*${id}: - capture group #1 followed by ${id} + :
([^,]*) - [capture group #2] - matches everything up to, but not including, the next comma (,)
(,.*)*$ - [capture group #3] - matches zero or more instances of comma followed by other characters to end of line
\2/p - print capture group #2
Alternatively, using a here-string to eliminate the pipe/sub-process call:
id='eu-central-1'
# desired id at start of input string:
sed -En "s/^(.*,)*${id}:([^,]*)(,.*)*$/\2/p" <<< 'eu-central-1:ami-bbbb,a-central-1:ami-aaaa,eu-north-1:ami-cccc,eu-west-1:ami-dddd'
# desired id at end of input string:
sed -En "s/^(.*,)*${id}:([^,]*)(,.*)*$/\2/p" <<< 'a-central-1:ami-aaaa,eu-north-1:ami-cccc,eu-west-1:ami-dddd,eu-central-1:ami-bbbb'
All three generate:
ami-bbbb

Answer 5

Defining , as line (record) separator and : as field separator, a simple condition over $1 prints the result.
echo -n a-central-1:ami-aaaa,eu-central-1:ami-bbbb,eu-north-1:ami-cccc,eu-west-1:ami-dddd |
awk 'BEGIN{RS=","; FS=":"}$1=="eu-central-1"{print $2}'