Bash/zsh: get substring from the last match in a string

In a shell script I have a variable $PATH which could look like this: /foo8232/test874-ab/bar/test-82-x. I now want to get a substring of this, getting as a result test-82-x in this specific case.

Thus I have to match the occurrences of /test.*, but only the last element OR from the end of the string. Whatever works best.

Any help on this?

Edit: Thanks for providing so many solutions so quickly. A few of them don't work though. Not sure, but maybe the problem is, that my path actually has whitespaces and more looks like this: /foo12323/test34 - 343bar/test - 234/test - 4323.txt. Sorry that I didn't mentioned it in the fi开发者_如何学JAVArst place, if that causes troubles.

For your specific path management, you've got the basename function :

basename /foo8232/test874-ab/bar/test-82-x
>>> test-82-x

or in a script :

#!/bin/bash
fullfilename=$(basename $1)
extension=${fullfilename##*.}
filename=${fullfilename%.*}

echo "fullfilename=$fullfilename"
echo "extension=$extension"
echo "filename=$filename"

which gives :

./file_name.sh /foo8232/test874-ab/bar/test-82-x
fullfilename=test-82-x
extension=
filename=test-82-x

The following should be POSIX-compliant

BASE=$(basename $PATH)
if echo $BASE | grep "^test.*" > /dev/null; then
  # do something awesome
fi

The last path element containing test could also be found with sed:

elem=$(echo $PATH | sed -r 's#.*/(test[^/]*).*#\1#')

Or without -r (using extended regular expressions):

elem=$(echo $PATH | sed 's#.*/\(test[^/]*\).*#\1#')

So you're looking for the last directory component that starts with test.

How about

IFS=/
for dir in $PATH; do
    case $dir in test*)
        match=$dir;;
    esac
done

echo pwd | awk -F \/ '{print $NF}'|grep test