unix - Variables as commands in bash scripts

Question

I am writing a very simple bash script that tars a given directory, encrypts the output of that, and then splits the resultant file into multiple smaller files since the backup media don't support huge files.

I don't have a lot of experience with bash scripting. I believe I'm having issues with quoting my variables properly to allow spaces in the parameters. The script follows:

#! /bin/bash

# This script tars the given directory, encrypts it, and transfers
# it to the given directory (likely a USB key).

if [ $# -ne 2 ]
then
    echo "Usage: `basename $0` DIRECTORY BACKUP_DIRECTORY"
    exit 1
fi

DIRECTORY=$1
BACKUP_DIRECTORY=$2
BACKUP_FILE="$BACKUP_DIRECTORY/`date +%Y-%m-%dT%H-%M-%S.backup`"

TAR_CMD="tar cv $DIRECTORY"
SPLIT_CMD="split -b 1024m - "$BACKUP_FILE""

ENCRYPT_CMD='openssl des3 -salt'

echo "$TAR_CMD | $ENCRYPT_CMD | $SPLIT_CMD"

$TAR_CMD | $ENCRYPT_CMD | $SPLIT_CMD 

say "Done backing up"

Running this command fails with:

split: "foo/2009-04-27T14-32-04.backup"aa: No such file or directory

I can fix it by removing the quotes around $BACKUP_FILE where I set $SPLIT_CMD. But, if I have a space in the name of my backup directory it doesn't work. Also, if I copy and paste the output from the "echo" command directly into the terminal it works fine. Clearly there's something I don't understand about how Bash is escaping things.

Answer 1

Simply don't put whole commands in variables. You'll get into a lot of trouble trying to recover quoted arguments.

Also:

1. Avoid using all-capitals variable names in scripts. Easy way to shoot yourself on the foot.
2. Don't use backquotes, use $(...) instead, it nests better.

---

#! /bin/bash

if [ $# -ne 2 ]
then
    echo "Usage: $(basename $0) DIRECTORY BACKUP_DIRECTORY"
    exit 1
fi

directory=$1
backup_directory=$2
current_date=$(date +%Y-%m-%dT%H-%M-%S)
backup_file="${backup_directory}/${current_date}.backup"

tar cv "$directory" | openssl des3 -salt | split -b 1024m - "$backup_file"