grep output from file

Question

How to remove the ### from grep output?

File containing:

### Invoked at: Wed Dec  7 22:24:35 2022 ###

My grep command:

grep -oP "Invoked at: \K(.*)" $file

Expected output:

Wed Dec  7 22:24:35 2022

Current output:

Wed Dec  7 22:24:35 2022 ###

Answer 1

Use this:

grep -oP "Invoked at: \K.*\d+" "$file"

(no need parenthesis).

The regular expression matches as follows:

Node	Explanation
Invoked at:	'Invoked at: '
\K	resets the start of the match (what is Kept) as a shorter alternative to using a look-behind assertion: look arounds and Support of K in regex
.*	any character except \n (0 or more times (matching the most amount possible))
\d+	digits (0-9) (1 or more times (matching the most amount possible))

Answer 2

You can remove a character from an output by adding | tr -d "#", so in your case this becomes:

grep -oP "Invoked at: \K(.*)" $file | tr -d "#"

Answer 3

You could match the last digit before matching any character except #

grep -oP "\bInvoked at: \K[^#]*\d" $file

Output

Wed Dec  7 22:24:35 2022

The pattern matches:

• \bInvoked at: \K Match Invoked at: and forget what is matched so far
• [^#]* Repeat 0+ times any char except #
• \d Match a digit

Regex demo

Or a more specific match for the whole example string using a case insensitive match with grep -oiP

^###\h+Invoked\h+at:\h+\K[a-z]+\h+[a-z]+\h+\d\d?\h+\d\d?:\d\d?:\d\d?\h+\d{4}(?=\h+###$)

Regex demo