Remove substring till first Token using regexp

I have the Path:开发者_Python百科

GarbageContainingSlashesAndDots/TOKEN/xyz/TOKEN/abc

How coukt I remove GarbageContainingSlashesAndDots? I know, it is before TOKEN, but Unfortunately, there are two substrings TOKEN in string. using sed s/.*TOKEN// makes my string to /abc, but I need /TOKEN/xyz/TOKEN/abc Thank You!!!

Divide and conquer:

$ echo 'Garbage.Containing/Slashes/And.Dots/TOKEN/xyz/TOKEN/abc' |
      sed -n 's|/TOKEN/|\n&|;s/.*\n//;p'
/TOKEN/xyz/TOKEN/abc

Is perl instead of sed allowed?

perl -pe 's!.*?(?=/TOKEN)!!'


echo 'GarbageContainingSlashesAndDots/TOKEN/xyz/TOKEN/abc' | perl -pe 's!.*?(?=/TOKEN)!!'
# returns:
/TOKEN/xyz/TOKEN/abc

Sed does not support non-greedy matching. Perl does.

I think you have bash, so it can be a simple as

$ s="GarbageContainingSlashesAndDots/TOKEN/xyz/TOKEN/abc"
$ echo ${s#*/}
TOKEN/xyz/TOKEN/abc

or if you have Ruby(1.9+)

echo $s | ruby -e 'print gets.split("/",2)[-1]'

Thank you for all suggestions, I've learnt something new. Finally I was able to reach my goal using grep -o

echo "GarbageContainingSlashesAndDots/TOKEN/xyz/TOKEN/abc" | grep -o "/TOKEN/.*/TOKEN/.*"

Using grep:

word='GarbageContainingSlashesAndDots/TOKEN/xyz/TOKEN/abc'
echo $word | grep -o '/.*'

echo "./a//...b/TOKEN/abc/TOKEN/xyz"|sed 's#.*\(/TOKEN/.*/TOKEN/.*\)#\1#'

UPDATE 2: have you tried this?

s!.*\(/TOKEN.+TOKEN.*\)!\1!

UPDATE: sorry, non-greedy matches are not supported by sed

Try this:

s/.*?TOKEN//

.*? matches only for the first occurance of TOKEN.