shell - Delete First and Second Period While Keeping Everything in Between (Linux) -

user calls script along input xx.xx.xx.xx x's numbers. i'm trying extract second set of xx left.

i've looked @ examples doing:

${1%.*}  <- deletes last . , else after ${1##*.} <- deletes last . , before

but without explanation on special characters do, i'm having trouble solving problem.

any appreciated.

using awk:

x='12.34.56.78' awk -f '.' '{print $2}' <<< "$x" 34

using pure bash (arrays):

n=$(ifs='.' read -ra arr <<< "$x" && echo "${arr[1]}") && echo "$n" 34

using pure bash without arrays 1:

y="${x#*.}" echo "${y%%.*}" 34

using pure bash without arrays 2:

n=$(ifs='.' && set -- $x && echo "$2") && echo "$n" 34

using sed:

sed 's/^[^.]*\.\([^.]*\).*$/\1/' <<< "$x" 34

Brazell