Grep characters before and after match?

Using this:

grep -A1 -B1 "test_pattern" file

will produce one line before and after the matched pattern in the file. Is there a way to display not lines but a specified number of characters?

The lines in my file are pretty big so I am not interested in printing the entire line but rather only observe the match in context. Any suggestions on how to do this?

Solution 1:

3 characters before and 4 characters after

$> echo "some123_string_and_another" | grep -o -P '.{0,3}string.{0,4}'
23_string_and

Solution 2:

grep -E -o ".{0,5}test_pattern.{0,5}" test.txt 

This will match up to 5 characters before and after your pattern. The -o switch tells grep to only show the match and -E to use an extended regular expression. Make sure to put the quotes around your expression, else it might be interpreted by the shell.

Solution 3:

You could use

awk '/test_pattern/ {
    match($0, /test_pattern/); print substr($0, RSTART - 10, RLENGTH + 20);
}' file

Solution 4:

You mean, like this:

grep -o '.\{0,20\}test_pattern.\{0,20\}' file

?

That will print up to twenty characters on either side of test_pattern. The \{0,20\} notation is like *, but specifies zero to twenty repetitions instead of zero or more.The -o says to show only the match itself, rather than the entire line.