How to output only captured groups with sed?

The key to getting this to work is to tell sed to exclude what you don't want to be output as well as specifying what you do want.

string='This is a sample 123 text and some 987 numbers'
echo "$string" | sed -rn 's/[^[:digit:]]*([[:digit:]]+)[^[:digit:]]+([[:digit:]]+)[^[:digit:]]*/\1 \2/p'

This says:

• don't default to printing each line (-n)
• exclude zero or more non-digits
• include one or more digits
• exclude one or more non-digits
• include one or more digits
• exclude zero or more non-digits
• print the substitution (p)

In general, in sed you capture groups using parentheses and output what you capture using a back reference:

echo "foobarbaz" | sed 's/^foo\(.*\)baz$/\1/'

will output "bar". If you use -r (-E for OS X) for extended regex, you don't need to escape the parentheses:

echo "foobarbaz" | sed -r 's/^foo(.*)baz$/\1/'

There can be up to 9 capture groups and their back references. The back references are numbered in the order the groups appear, but they can be used in any order and can be repeated:

echo "foobarbaz" | sed -r 's/^foo(.*)b(.)z$/\2 \1 \2/'

outputs "a bar a".

If you have GNU grep (it may also work in BSD, including OS X):

echo "$string" | grep -Po '\d+'

or variations such as:

echo "$string" | grep -Po '(?<=\D )(\d+)'

The -P option enables Perl Compatible Regular Expressions. See man 3 pcrepattern or man 3 pcresyntax.

Sed has up to nine remembered patterns but you need to use escaped parentheses to remember portions of the regular expression.

See here for examples and more detail

you can use grep

grep -Eow "[0-9]+" file