CODEWITHSUNDEEP
bash
viswa
viswa

Reputation: 51

shell script backing up the source directory to backup directory

#!/bin/bash

BACKUP=backup_date
SOURCE=Arithmetic_Operators

echo "Taking backup from ${SOURCE} directory to backup directory ${BACKUP} .."

# Checking the source directory ${SOURCE} exists or not ! if not exists die
# Script is unsuccessful with exit status # 1

[ ! -d $SOURCE ] && echo "source directory $SOURCE not found"
exit 1

# Checking the backup directory ${BACKUP} exists or not ! if not exists die
# Script is unsuccessful with exit status # 2

[ ! -d $BACKUP ] && echo "backup directory $BACKUP not found"
exit 2

# Let Start the backing up

tar cvf $SOURCE $BACKUP 2> /wrong/logs.txt

if [ $? -ne 0 ]
then
   # die with unsuccessful shell script termination exit status # 3
   echo "An error occurred while making a source directory backup, see /wrong/logs.txt file".
   exit 3 
fi

This is my script to backing up the source directory(Arithmetic_Operators) to destination directory (backup_date), while running the script my script was ending with message

Taking backup from Arithmetic_Operators directory to backup directory backup_date ..
source directory Arithmetic_Operators not found

where i did mistake why this script is not running could you please help me on this ?

Upvotes: 0

Views: 134

Answers (1)

Jonathan Leffler
Jonathan Leffler

Reputation: 753870

These lines mean the script exits unconditionally:

[ ! -d $SOURCE ] && echo "source directory $SOURCE not found"
exit 1

You probably meant:

if [ ! -d $SOURCE ]
then
    echo "source directory $SOURCE not found" >&2
    exit 1
fi

Or maybe:

[ ! -d $SOURCE ] && { echo "source directory $SOURCE not found" >&2; exit 1; }

Note that error messages should be sent to standard error, not standard output. It wouldn't be a bad idea to include the name of the script ($0) in the messages; it helps identify which script generated/detected the problem.

You have a similar problem after checking the $BACKUP directory.

Also, as a general rule, enclose variable references in double quotes:

[ ! -d "$SOURCE" ] && { echo "source directory $SOURCE not found" >&2; exit 1; }

(The second reference is already inside double quotes, of course.)

Upvotes: 1

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