Get last line of shell output as a variable

giopas picture giopas · Jul 13, 2015 · Viewed 68.8k times · Source

I am working on a shell script with exiftool to automatically change some exif tags on pictures contained in a certain folder and I would like to use the output to get a notification on my NAS (a QNAP) when the job is completed.

Everything works already, but - as the notification system truncates the message - I would like to receive just the information I need, i.e. the last line of the shell output, which is for example the following:

Warning: [minor] Entries in IFD0 were out of sequence. Fixed. - 2015-07-12 15.41.06.jpg                                                                             
 4512 files failed condition                                                      
  177 image files updated

The problem is that currently I only receive the following notification:

Exiftool cronjob completed on Camera: 4512 files failed condition

What I would like to get instead is:

Exiftool cronjob completed on Camera: 177 image files updated

The script is the following:

#!/bin/sh
# exiftool script for 2002 problem
dir="/share/Multimedia/Camera"
cd "$dir"
FOLDER="$(printf '%s\n' "${PWD##*/}")"
OUTPUT="$(exiftool -overwrite_original -r '-CreateDate<DateTimeOriginal' -if '$CreateDate eq "2002:12:08 12:00:00"' -if '$DateTimeOriginal ne $CreateDate' *.[Jj][Pp][Gg])"
/sbin/notice_log_tool -a "Exiftool cronjob completed on ${FOLDER}: ${OUTPUT}" --severity=5
exit 0

To do that I played with the $OUTPUT variable using | tail -1, but probably I make some basic errors and I receive something like:

Exiftool cronjob completed on Camera: 4512 files failed condition | tail -1

How to do it in the right way? Thanks

Answer

user464502 picture user464502 · Jul 13, 2015

Put the tail inside the capturing parens.

OUTPUT=$(exif ... | tail -1)

You don't need the double quotes here. I'm guessing that you tried

OUTPUT="$(exif ...) | tail -1"