Question

Is there a way to tell sed to output only captured groups?

For example, given the input:

This is a sample 123 text and some 987 numbers

And pattern:

/([\d]+)/

Could I get only 123 and 987 output in the way formatted by back references?

Answer 1

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. This technique depends on knowing how many matches you’re looking for. The grep command below works for an unspecified number of matches.

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:

• use extended regular expressions (-r)
• 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) (on one line)

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:

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

It may also work in BSD, including OS X:

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

These commands will match any number of digit sequences. The output will be on multiple lines.

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.