Replacing variable with command line argument in bash script

Question

I would like to call the script ./archive-it.sh with folder name ( like project1 ) without trailing / and expect to archive it, I am sure its pretty simple, as I am new to bash script i am stuck. Can someone help.

but currently i am getting error here

$target=$(echo "$1"-`date +%m-%d-%Y-%H_%M`)

Here is the script

!/bin/bash                        
######### archive-it.sh ############

echo "Argument  $1";
$target=$(echo "$1"-`date +%m-%d-%Y-%H_%M`)
tar -czvf $target.tar.gz  $1  

echo "####################################"
echo "ARCHIVE Successfully done at :"
echo "${target}" ;

also suggest me if its good to exclude .git folder, as I am taking only backup on a daily basis.

naab · Accepted Answer · 2014-04-12 18:03:12Z

3

To assign a variable, it's target="something", then to access it you use $target

Also add quotes around $1 if the path contains spaces

echo "Argument  $1";

if [[ ! -d "${1}" ]]; then
    echo "Folder '$1' doesn't exists"
    exit 1
fi

#filter git files if you want
nogitfiles="$(find "${1}" -name * | grep -v .git)"

target="$(date +"$1-%m-%d-%Y-%H_%M")"
tar -czvf "${target}.tar.gz"  ${nogitfiles}  

echo "####################################"
echo "ARCHIVE Successfully done at :"
echo "${target}" ;

edited Apr 12, 2014 at 18:03

answered Apr 12, 2014 at 10:03

naab

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perreal · Accepted Answer · 2014-04-12 10:15:46Z

1

You don't need the echo at all:

target="$1$(date +%m-%d-%Y-%H_%M)"

answered Apr 12, 2014 at 10:15

perreal

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Comments

chepner · Accepted Answer · 2014-04-12 12:45:11Z

0

In addition to assigning to target, not $target, and instead of concatenating $1 and the output of date, you can include $1 in the format string for date:

target=$(date +"$1-%m-%d-%Y-%H_%M")

answered Apr 12, 2014 at 12:45

chepner

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