[SOLVED] Remove extra dot in a String by grep/awk/cut

Issue

I have the following text in a file, for example, output.txt

[test.tracking_utils] INFO: Tracking subtool usage: main_test
[TEST & SPEC] INFO: Uploaded file test.zip with bucket URI test/20210804144418.zip.

How to use grep o something similar to get the value as test/20210804144418.zip?

I tried

tail output.txt | grep ". test/" | awk {print $NF}
tail output.txt | grep -m1 .test/ | rev | cut -d' ' -f1 | rev

it returns test/20210804144418.zip. with extra . do you know how to remove the extra . at the end? any suggestion?

Solution

You may use this grep:

grep -oE -m1 'test/[^[:blank:]]+\.[^.]+' file

test/20210804144418.zip

RegEx Details:

• test/: Match test/
• [^[:blank:]]+: Match 1+ of non-whitespace characters
• \.: Match a dot
• [^.]+: Match 1+ of any non-dot characters

Answered By - anubhava

Answer Checked By - Willingham (WPSolving Volunteer)