Can you grep a file using a regular expression and only output the matching part of a line?

Question

I have a log file which contains a number of error lines, such as:

Failed to add [email protected] to database

I can filter these lines with a single grep call:

grep -E 'Failed to add (.*) to database'

This works fine, but what I'd really like to do is have grep (or another Unix command I pass the output into) only output the email address part of the matched line.

Is this possible?

Answer 1

You can use sed:

grep -E 'Failed to add (.*) to database'| sed 's/'Failed to add \(.*\) to database'/\1'

Answer 2

or python:

cat file | python -c "import re, sys; print '\r\n'.join(re.findall('add (.*?) to', sys.stdin.read()))"

Answer 3

sed is fine without grep:

sed -n 's/Failed to add \(.*\) to database/\1/p' filename

Answer 4

This should do the job:

grep -x -e '(?<=Failed to add ).+?(?= to database)'

It uses a positive look-ahead assertion, followed by the match for the email address, followed by a postivie look-behind assertion. This insures that it matches the entire line, but only actually consumes (and thus returns) the email address part.

The -x option specifies that grep should match lines rather than the whole text.

Answer 5

If you want to use grep, it would be more appropiate to use egrep;

About egrep

Search a file for a pattern using full regular expressions.

grep will not always have as complete of functionality for regex.

Answer 6

Recent versions of GNU grep have a -o option which does exactly what you want. ( -o is for --only-matching).