bash: extract only the part of the string following the dot

Question

Say I have a string such as

="/home/Rob/.SpaceVim"

and I want to extract the string following the dot. How do I do this?

I know $ is also used to represent the last character, and \ can be used to make character in string literal. So I've tried stuff like awk '\.$' but not had any success.

Answer 1

Could be:

echo "/home/Rob/.SpaceVim" | grep -o "\..*$"

or

echo "/home/Rob/.SpaceVim" | sed 's;[^.]*;;'

---

Without dot and GNU grep

echo "/home/Rob/.SpaceVim" | grep -Po "(?<=\.).*$"

or sed

echo "/home/Rob/.SpaceVim" | sed 's;^.*\.;;'

you could even include cut

echo "/home/Rob/.SpaceVim" | cut -d'.' -f2-

(beware you'll see variations in behaviour among those for lines that contain no dot or more than one dot, but you've not specified what to do in that case).

Answer 2

Read man bash.

a="/home/Rob/.SpaceVim"
echo ${a##.*/}