Using sed and regex to grab the last part of a url

Question

I am trying to get sed to match the last part of the url and output exactly that. For instance:

echo "http://randomurl/suburl/file.mp3" | sed (expression)

should produce the result:

file.mp3

So far I have tried sed 's|\([^/]+mp3\)$|\1|g', but it just displays the whole URL. Maybe there is something I don’t see here, but in any case, help would be greatly appreciated!

Answer 1

it works:

 echo "http://randomurl/suburl/file.mp3" | sed 's#.*/##'

Answer 2

basename - your good friend.

> basename "http://randomurl/suburl/file.mp3"
=> file.mp3

Answer 3

:

$ echo "http://randomurl/suburl/file.mp3" | sed -r 's|.*/(.*)$|\1|'
file.mp3

:

• | / s.
• , /.

: bash :

$ url="http://randomurl/suburl/file.mp3"
$ echo ${url##*/}
file.mp3

Answer 4

echo 'http://randomurl/suburl/file.mp3' | grep -oP '[^/\n]+$'

, grep.