print everything after a slash

Question

I found how to print everything before a slash, however I need to print everything after a slash. I have a string like blablabal/important and I need to print only important. How should I modify the line below to get printed everything after a slash and not before?

sed 's:/[^/]*$::'

Answer 1

Just delete everything until the last slash:

$ echo "blablabal/important" | sed 's:.*/::'
important

It also works with multiple slashes:

$ echo "blablabal/not/ver/interesting/important" | sed 's:.*/::'
important

Answer 2

It looks like the other answers are assuming that you have multiple lines of data in a file, which are all to be processed. In this case sed or awk (or possibly cut) would be the best tools for processing all the lines in one go.

However, if you just have one line in a shell variable (assuming you're using bash), you can use shell expansions to achieve the desired result, without having to spawn utilities in external processes:

$ var="interesting/bla/blablabal/important"
$ 
$ # everything after the first slash:
$ echo "${var#*/}"
bla/blablabal/important
$ # everything after the last slash:
$ echo "${var##*/}"
important
$ # everything before the first slash:
$ echo "${var%%/*}"
interesting
$ # everything before the last slash:
$ echo "${var%/*}"
interesting/bla/blablabal
$ 

---

Alternatively, I assume your slash-separated strings are file paths. If that is the case, you can use dirname and basename to get the path and filename components:

$ # everything before the last slash:
$ dirname "$var"
interesting/bla/blablabal
$ # everything after the last slash:
$ basename "$var"
important
$ 

Answer 3

Since you mentioned in your comment that the file only has one slash, why not just use awk?

Example:

 ❱ echo "blablabal/important" | awk -F'/' '{print $1}'
blablabal
 ❱ echo "blablabal/important" | awk -F'/' '{print $2}'
important

Answer 4

This sould work:

echo "blablabal/important" | sed 's:[^/]*/\(.*\):\1:'

The first part [^/]*, matches everything BUT a slash, then we have a literal slash /, and a "matches everything" .* inside escaped parenthesis \(.*\). The escaped parenthesis are there to "save" the matching result, meaning that it will contain everything after the first slash in this case. The \1 refer to the first matched group, in this case we only have one group, because there is only one \(...\) pair.

In sed, the character you type after the s will be the delimiter. It is a common practice to use / as delimiter, but is this case it is easier to use something else, although it would be possible to use a slash too.

Answer 5

sed -et -e's|/|&\n|;//D' <in >out

...will print everything after the first slash on any line which contains one, or else print any other unmodified. If you drop the // it will only print lines it modifies.

You may need to use a literal \newline in place of the n in the \n escape though, depending on your sed version.

Also, if you specify an occurrence count...

sed -et -e's|/|&\n|num;/\n/D'

...you can delete up to the numth slash on a line without affecting any line which doesn't contain at least num slashes.

Answer 6

grep solution:

$ echo "blablabal/important" | grep -oP "/\K.*"
important

\K drops the matched string in output