3

Executing the following command works as expected:

find /path/to/logs/ \( -type f -name '*20161005.log' \) -print0 | tar -czvf /path/to/backups/backup_logs_20161005.tar.gz --null -T -

But, trying to use this within a bash script, I get an empty .tar.gz file:

#!/bin/bash
now="$(date)"
printf "Current date and time: $now"

## working paths
LOGSPATH="/path/to/logs/"
BACKPATH="/path/to/backups/"

## date to work with
DATETIME=`date -d "yesterday 13:00 " '+%Y%m%d'`

## find files matching the working date and .tar.gz the result
printf "\nStarting yesterday logs backup, please wait..."
find "$LOGSPATH" \( -type f -name '*"$DATETIME".log' \) -print0 | tar -czvf "$BACKPATH"backup_logs_"$DATETIME".tar.gz --null -T -
printf "\ndone!"

## delete backup files
printf "\nRemoving logs from yesterday..."
find "$LOGSPATH" -type f -name '*"$DATETIME".log' -delete
printf "\ndone!\n"

Output:

Current date and time: Thu Oct 6 16:14:29 WEST 2016
Starting yesterday logs backup, please wait...
done!
Removing logs from yesterday...
done!

What I am missing when placing this command within a bash script?

Zuul
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1 Answers1

3

Replace 

-name '*"$DATETIME".log'

with

-name "*$DATETIME.log"

The single quotes are very strong, the double quotes allow $VARIABLE inside.

kubanczyk
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