sed - how to do regex groups using sed

Question

Is there anyway you can do regex match group using sed like java regex pattern/match/group?

if i have string like

test-artifact-201251-balbal-0.1-SNAPSHOT.jar

how do I use sed just to get the result like:

test-artifact-0.1-SNASHOT.jar

I am wondering does sed allow you to do something like java regex, you define the pattern like:

([a-z]*-[a-z]*-)([0-9]*-)([a-z]*-)([.]*SNAPSHOT.jar)

and then you can get the results as an array like:

test-artifact-
201251-
balbal-
0.1-SNAPSHOT.jar

Answer 1

You have to escape parentheses to group expressions:

\([a-z]*-[a-z]*-\)\([0-9]*-\)\([a-z]*-\)\([.]*SNAPSHOT.jar\)

And use them with \1, \2, etc.

---

EDIT: Also note just before SNAPSHOT that [.] will not match. Inside brackets . is literal. It should be [0-9.-]*

Answer 2

This is what Birei and Thor mean:

sed -r "s/([a-z]*-[a-z]*-)([0-9]*-)([a-z]*-)(.*)/\1\n\2\n\3\n\4/"

Output:

test-artifact-
201251-
balbal-
0.1-SNAPSHOT.jar

Answer 3

infact for those regular string, awk could save you from grouping. :)

you just give the part index number you want:

awk 'BEGIN{FS=OFS="-"}{print $1,$2,$5,$6}' 

output:

kent$  echo "test-artifact-201251-balbal-0.1-SNAPSHOT.jar"|awk 'BEGIN{FS="-";OFS="-"}{print $1,$2,$5,$6}'
test-artifact-0.1-SNAPSHOT.jar

Answer 4

If you are searching for an easier way I guess this might be of your help! :)

echo "est-artifact-201251-balbal-0.1-SNAPSHOT.jar" | cut -d- -f1,2,5,6

"-" used as delimeter and fields 1,2,5,6 are printed.

Note: This would require you to know the exact position of the field.