Pass dynamically generated parameters to command inside script

Question

I have a script which calls the rsync command with some dynamically generated parameters but I'm having trouble passing them correctly.

Here's some excerpt:

logfile="$logDir/$(timestamp) $name.log"
echo "something" >> "$logfile"

params="-aAXz --stats -h --delete --exclude-from $exclude --log-file=$logfile $src $dest"
if [ "$silent" = "" ]; then
    params="-v $params --info=progress2"
fi
rsync $params

If the logfile is e.g. /tmp/150507 test.log, the something statement is actually written to /tmp/150507 test.log, but rsync writes its logs to /tmp/150507 (everything after the first blank removed).

If I explicitly quote the name of the logfile in the params, rsync throws an exception:

params="-aAXz --stats -h --delete --exclude-from $exclude --log-file=\"$logfile\" $src $dest"

The error:

rsync: failed to open log-file "/tmp/150507: No such file or directory (2)
Ignoring "log file" setting.

How can I generate the params dynamically without losing the ability to use blanks in the filenames?

anubhava · Accepted Answer · 2022-03-16 21:00:40Z

5

More quoting needed around log file name:

declare -a params

params=(-aAXz --stats -h --delete --exclude-from "$exclude" --log-file="$logfile" "$src" "$dest")

if [[ "$silent" = "" ]]; then
    params=( -v "${params[@]}" --info=progress2 )
fi

rsync "${params[@]}"

This is the case where you should consider using BASH arrays to constitute a dynamic command line.

edited Mar 16, 2022 at 21:00

answered May 7, 2015 at 10:53

anubhava

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4 Comments

suamikim Over a year ago

Ain't that exactly what I mentioned in the 2nd half of my post (starting with "If I explicitly quote...") which leads to the rsync error?

Charles Duffy Over a year ago

@suamikim, no, because in that 2nd half of your post you're creating a string; in the answer here you're creating an array.

Charles Duffy Over a year ago

@anubhava, ...though the -v "$params" is a bit odd; you're adding the first element of the array in directly after the -v? I think a more accurate representation would be params=( -v "${params[@]}" --info=progress2 ); it's not += because the OP is replacing the prior first argument with -v.

anubhava Over a year ago

Thanks @CharlesDuffy for your insight. It is corrected now

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