Not operator in regex

Question

I have a file and I wish to grep out all the lines that do not start with a timestamp. I tried using the following regex but it did not work:

cat myFile | grep '^(?!\[0-9\]$).*$'

Any other suggestions or something that I might be doing wrong here?

It belongs to http://unix.stackexchange.com/, but can't find it between the off topic options. Voted for http://superuser.com/ — BlackBear, Feb 06 '12 at 19:19

score 8 · Accepted Answer · answered Feb 06 '12 at 19:23

8

Why not simply use grep -v option like this to negate:

grep -v "<pattern>" file

Let's say you want to grep all the lines in a shell script that are not commented ( do not have # at start ) then you can use:

grep -v "^\s*#" file.sh

answered Feb 06 '12 at 19:23

anubhava

1

The only *right* answer here: Don't try to negate the pattern, negate the match result. +1 – Platinum Azure Feb 06 '12 at 19:25

score 0 · Answer 2 · answered Feb 06 '12 at 19:18

0

You don't need a not operator, just use grep as it is most easily used: finding a pattern:

grep '^[0-9]' myFile

answered Feb 06 '12 at 19:18

beerbajay

score 0 · Answer 3 · answered Feb 06 '12 at 19:18

Try this:

cat myFile | grep '^\d\d\d\d-\d\d-\d\d \d\d:\d\d:\d\d'

This assumes your timestamp is of the pattern dddd-dd-dd dd:dd:dd, but you change it to what matches your timestamp if it's something else.

Note: Unless you're using some kind of cmd chaining, grep pattern file is a simpler syntax

BTW: Your use of a double-negative makes me unsure if you want the timestamp lines or you want the non-timestamp lines.

3 Answers3