Grep characters before and after match?

2018-06-06 07:28:43

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?

之前3个字符和之后4个字符

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

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.

你可以使用

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

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