bash - Grep characters before and after match?

Question

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?

Answer 1

3 characters before and 4 characters after

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