Using sed for extracting substring from string

Question

I just started using sed from doing regex. I wanted to extract XXXXXX from *****/XXXXXX> so I was following

sed -n "/^/*/(\S*\).>$/p"

If I do so I get following error

sed: 1: "/^/ /(\\S ).>$/p": invalid command code *

I am not sure what am I missing here.

Answer 1

Try:

$ echo '*****/XXXXXX>' | sed 's|.*/||; s|>.*||'
XXXXXX

The substitute command s|.*/|| removes everything up to the last / in the string. The substitute command s|>.*|| removes everything from the first > in the string that remains to the end of the line.

Or:

$ echo '*****/XXXXXX>' | sed -E 's|.*/(.*)>|\1|'
XXXXXX

The substitute command s|.*/(.*)>|\\1| captures whatever is between the last / and the last > and saves it in group 1. That is then replaced with group 1, \\1 .

Answer 2

In my opinion awk performs better this task. Using -F you can use multiple delimiters such as "/" and ">":

echo "*****/XXXXXX>" | awk -F'/|>' '{print $1}'

Of course you could use sed , but it's more complicated to understand. First I'm removing the first part (delimited by "/") and after the second one (delimited by ">"):

echo "*****/XXXXXX>" | sed -e s/.*[/]// -e s/\>//

Both will bring the expected result: XXXXXX.

Answer 3

with grep if you have pcre option

$ echo '*****/XXXXXX>' | grep -oP '/\K[^>]+'
XXXXXX

• /\\K positive lookbehind / - not part of output
• [^>]+ characters other than >

Answer 4

echo '*****/XXXXXX>' |sed 's/^.*\/\|>$//g'
XXXXXX

Start from start of the line, then proceed till lask / ALSO find > followed by EOL , if any of these found then replace it with blank.