How to pass variable number of parameters to a shell command?

Question

I am writing a script for backup like this:

backup.sh:

dir="$1"
mode="$2"
delta="$3"

for file in "$dir/backup."*".$mode.tar.gz"; do
    [ "$file" -nt "$ref" ] && ref="$file"
done

if [ "$delta" = "true" ]; then
    delta_cmd=-N "'$ref'"
fi

backup_file="$dir/backup.$(date +%Y%m%d-%H%M%S).$mode.tar.gz"

case "$mode" in
    config)
        tar -cpzvf "$backup_file" $delta_cmd \
            /etc \
            /usr/local
            ;;
    # still other modes here...
esac

I want to pass a single variable $delta_cmd to the tar command so that it tars all files or only delta files since last backup depending on the value of $delta.

The above code creates an error message and does not tar the delta files correctly if $delta is set to true. How to fix it?

P.S: The script would better be POSIX compatible.

It's possible to make this POSIX compatible, but only by either introducing security vulnerabilities (using eval) or overwriting $@ (which is quite not compatible with the use of a delta_cmd string) -- which would mean encapsulating the code in a function if you don't want to override the script-global value for that array. See BashFAQ #50 for a description of why string variables (as opposed to arrays) can't be safely used to store argument lists or commands, at mywiki.wooledge.org/BashFAQ/050 — Charles Duffy
– Charles Duffy, Commented Aug 1, 2016 at 19:05
Could you explain a bit more about the overwriting $@ way? — Danny Lin
– Danny Lin, Commented Aug 1, 2016 at 19:19
Added an answer. BTW, there's quite a lot in this code that could afford to be cut down to create a proper MCVE (minimal, complete, verifiable example; see stackoverflow.com/help/mcve) -- I'm very much hesitant to include content in my answer that includes such bad practices as parsing ls, but a better-practices replacement for that code is also somewhat outside the scope of the question at hand. — Charles Duffy
– Charles Duffy, Commented Aug 1, 2016 at 19:23
...that said, please see BashFAQ #3 at mywiki.wooledge.org/BashFAQ/003 for guidance on a best-practices way to find the newest, or oldest, file in a directory. (We also have answers in the StackOverflow knowledge base about using GNU find to efficiently and -- unlike ls -- robustly find the most or least recent N files). — Charles Duffy
– Charles Duffy, Commented Aug 1, 2016 at 19:24

Charles Duffy · Accepted Answer · 2016-08-09 12:27:24Z

1

As a POSIX-compliant approach, consider:

set --                  # clear $@
if [ -f "$ref" ]; then
  set -- "$@" -N "$ref" # add -N "$ref" to $@
fi

tar ... "$@" ...        # expand $@ into command line

To put this all in context, and protect the main argument list against overwrite, might look like:

#!/bin/sh

main() {
    # if current shell supports "local", prevent variables from leaking
    # ...some "POSIX" shells, such as ash, will be fine with this.
    local dir mode delta target_file backup_file 2>&1 ||:

    dir=$1
    mode=$2
    delta=$3

    set -- # clear $@

    for file in "$dir/backup."*".$mode.tar.gz"; do
        [ "$file" -nt "$ref" ] && ref="$file"
    done

    if [ "$delta" = "true" ]; then
        set -- "$@" -N "$ref"
    fi

    target_file="$dir/backup.$(date +%Y%m%d-%H%M%S).$mode.tar.gz"

    case "$mode" in
        config)
            tar -cpzvf "$target_file" "$@" \
                /etc \
                /usr/local
            ;;
        # still other modes here...
    esac
}

main "$@"

edited Aug 9, 2016 at 12:27

answered Aug 1, 2016 at 19:21

Charles Duffy

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3 Comments

Danny Lin Over a year ago

Thank you. This is helpful. If we put the code fragment to a function in sub-shell such as backup () ( ... ) it would not overwrite $@ would it?

Charles Duffy Over a year ago

correct, it would only overwrite the $@ for that function. You actually don't need backup() ( ... ), backup() { ... } would suffice, since a function is in a different stack frame and so has its own argument list.

Danny Lin Over a year ago

Note that there had been an error in the original post which was corrected: target_file should be backup_file.

anubhava · Accepted Answer · 2016-08-01 16:55:09Z

1

You should be using BASH arrays for storing partial/full command lines:

#!/bin/bash

DIR=/home/sysop/backup
mode=main
delta=false

REF=$(ls -t "$DIR"/system.*.$mode.tar.gz "$DIR"/system.*.$mode-delta.tar.gz 2>/dev/null | head -n 1)
REF=$(readlink -f "$REF")

if [ "$delta" = true ]; then
    delta_cmd=(-N "$REF")
    delta_suffix=("-delta")
fi

target_file="$DIR/system.$(date +%Y%m%d-%H%M%S).$mode$delta_suffix.tar.gz"

tar -cpzvf "$target_file" "${delta_cmd[@]}" \
    /etc \
    /usr/local \
    /var/log \
    /var/spool \
    /home/*/logs

I would also suggest avoiding parsing of ls command's output in your script.

edited Aug 1, 2016 at 16:55

answered Aug 1, 2016 at 16:43

anubhava

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11 Comments

Danny Lin Over a year ago

It does not seem to work POSIX-ly due to syntax error. I also tried using bash, but the result is error as previously if $delta=true.

anubhava Over a year ago

This script should run on BASH. Can you tell me what is exact error while using BASH?

anubhava Over a year ago

Oh you have misplaced quoted in array assignment. Note that I have delta_cmd=(-N "$REF") but you are showing delta_cmd=(-N "'$ref'")

anubhava Over a year ago

You are not copying correctly from my answer. I have "${delta_cmd[@]}" but you have $delta_cmd.

Danny Lin Over a year ago

You are right. After replacing $delta_cmd to "${delta_cmd[@]}", it works now.

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