Creating a zip file with a timestamp

Question

I have been trying to create a script which compresses a directory with a timestamp and move it to another location on the same server, but I cant seem to get TESTDATE to display on the zip file, resulting in filed name .zip.

#!/bin/sh

# date
TESTDATE="date +%b-%d-%y"

# dirs to backups
DIRS="/volume1/NetBackup"

# Save zip file elsewhere on the server
DEST="/volume1/NetBackup/Backup"

# set to "no" keep old zip file and add new file
DELETE_OLD_ZIP_FILES="no"

#Path to binary files
BASENAME=/usr/bin/basename
ZIP=/usr/syno/bin/zip

for d in $DIRS
do
echo $DIRS
    zipfile="${DEST}/$(${TESTDATE} ${d} ${BASENAME}).zip"
    echo -n "Creating $zipfile..."
            # create zip file
                ${ZIP} -r $zipfile $d &>/dev/null && echo "Done!"
                done

Solution

You have various problems there.

• First, you want to assign the result of the command date +%b-%d-%y to the TESTDATE variable. To do so, you need to use command substitution :

  TESTDATE=$(date  +%b-%d-%y)
  

  What your script is currently doing is assignig the string date +%b-%d-%y to TESTDATE, not the result of the command. You can test by running this:

  TESTDATE="date +%b-%d-%y"
  echo $TESTDATE 
  

  which will print date +%b-%d-%y.

• Second, you are using command substitution where you shouldn't, what you want is:

    zipfile="${DEST}/${TESTDATE} ${d} ${BASENAME}.zip"
  

  So, you don't want the parentheses. I also have no idea what you are trying to do with basename there. basename is a default program and you don't need to place it in a variable to use it. The command above sets zipfile to /volume1/NetBackup/Backup/${TESTDATE} ${d} /usr/bin/basename which is not a valid file name. If you explain what you're trying to do with basename I might be able to help but as it is it makes no sense.

• You seem to think that DIRS is a list, it's not, it is just the single directory /volume1/NetBackup.

• Another point is a matter of style. You should always avoid using CAPITAL variable names because environmental variables are all caps and this might lead to problems if you use the same name.

Anyway, if I understand what you're trying to do here, you could write the script like this (I have made dirs into an array since I think that's what you wanted):

#!/usr/bin/env bash

# date
testdate=$(date +%b-%d-%y)

# dirs to backup
dirs=("/volume1/NetBackup" "/volume1/foobar")

# Save zip file elsewhere on the server
dest="/volume1/NetBackup/Backup"

# set to "no" keep old zip file and add new file
delete_old_zip_files="no"

#Path to binary files
zip=/usr/syno/bin/zip

base=$(/usr/bin/basename ${dest})
echo "bb $base";

for d in ${dirs[@]}
do
    ## I assume this is what you wanted to do, it will just
    ## return the name of the current $d directory
    base=$(/usr/bin/basename ${d})
    zipfile="${dest}/${testdate} ${base}.zip"
    echo -n "Creating $d $zipfile..."
    # create zip file
    ${zip} -r $zipfile $d  && echo "Done!" || echo ""
done

The example script above would create /volume1/NetBackup/Backup/Mar-05-14 NetBackup.zip and /volume1/NetBackup/Backup/Mar-05-14 foobar.zip. By the way, I would also strongly suggest you avoid creating file names with spaces, you'll thank me later.