How to print before and after dot text Unix - shell

Question

I am trying to print only specific output from sentence like below
Before and after dot text should be printed
InputVar="ABC SDFSG XYZ.AFGAJK JKK"
Expected output :
XYZ.AFGAJK
I am using cut command not working
echo "$InputVar" | cut -d'' -f2
Any other approach ?

Answer 1

Here are a few suggestions. awk with RS set to a space seems easiest. YMMV
$ echo "$InputVar" | cut -d ' ' -f 3
XYZ.AFGAJK
$ echo "$InputVar" | awk '/\./' RS=' '
XYZ.AFGAJK
$ echo "$InputVar" | awk '{for(i=1;i<=NF;i++) if(match($i,"\\.")) print $i}'
XYZ.AFGAJK
$ echo "$InputVar" | sed -n 's/.* \([^ .]*[.][^ .]*\) .*/\1/p'
XYZ.AFGAJK

Answer 2

Using cut:
If you really want to use cut, then you could try:
echo "$InputVar" | cut -d' ' -f3
Which uses a space character as a delimiter (you originally had an empty string, which is not allowed), and extracts field 3 rather than field 2.
Using grep:
You can use grep rather than cut, to match & extract specifically what you want:
echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+'
Explanation:
The -E option is for extended regex
The -o option is for extracting the matched component only
The regex matches a literal ., surrounded by a non-empty sequence of non-space characters
Comparing the two methods:
Either of these will work with your shown example. But, suppose the input string was instead:
InputVar="ABC SDFSG XYZ.AFGAJK JKK XYZ.ABC"
The version using grep would give all the matches (a literal . with non-space characters on either side).
Using cut however, you would need to specify the specific fields you want, i.e.
$ echo "$InputVar" | cut -d' ' -f3,5
XYZ.AFGAJK
XYZ.ABC
If you instead wanted just the n-th match, using the grep approach, you could use sed to select the n-th match, e.g.
$ echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+'
XYZ.AFGAJK
XYZ.ABC
$ echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+' | sed '1q;d'
XYZ.AFGAJK
$ echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+' | sed '2q;d'
XYZ.ABC