Unable to execute bash script through cron/crontab

Question

I have the below cmd in extract.sh .

It works when execute in terminal to search for "ABC123" in orig.txt and output to result.txt the remaining content after "ABC123" is found.

Unfortunately when I run the script via cron/crontab, it seem to ignore/skip the "more" command and just output orig.txt to result.txt.

Any help is appreciated.

---------------------extract.sh-----------------------------------------

#!/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin
. $HOME/.profile
more +/"ABC123" orig.txt > result.txt

---

----------------------crontab-------------------------------------------

SHELL=/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin
* * * * * /root/extractpdf_v3.sh
* * * * * /bin/bash /root/extract.sh

---

Answer 1

You can use sed instead of more, as you mention in you comment you need the content of the file after the line that match specific word. For example input file:

Test before
Test match
Test after

command:

sed -n -e '/match/,$p'

output:

Test match
Test after

• -n disables default behaviour of sed of printing each line after executing its script on it,
• -e indicated a script to sed,
• /match/,$ is an address (line) range selection meaning the first line matching the 'match' regular expression (like grep) to the end of the file ($)
• p is the print command which prints the current line.

For more details, the original answer https://stackoverflow.com/questions/7103531/how-to-get-the-part-of-file-after-the-line-that-matches-grep-expression-first