I'm newbie in bash scripting

I want to format command with printf to prepare execution and assign into variable.

But it is executing immediately. It is listing /opt/cinar/packages/libcnrhttp2_* instead of writing it as string into command string.

I was expecting this:

ssh [email protected] ls /opt/cinar/packages/libcnrhttp2_* -lr | awk 'NR==1'

enter image description here

declare -r libcnrhttp2="libcnrhttp2_"
declare -r root="/opt/cinar/packages"
declare -r destination="/opt/cinar/packages"
declare -r id=ctopkaya
declare -r host=
declare -r remote=$id"@"$host

declare cmd=""
printf -v cmd ssh %s ls %s/%s* -lr | awk 'NR==1' $remote /opt/cinar/packages libcnrhttp2_

cd $destination
echo "--------------"
echo $cmd
echo "--------------"
  • 3
    Please copy and paste text from your terminal rather than posting screenshots.
    – user413007
    Jul 22, 2020 at 8:02
  • It's unclear what you want to happen. You print the string saved in $cmd by printf -v cmd using echo $cmd. This is why that string is outputted. The generated command in $cmd is not executed by the script that you show.
    – Kusalananda
    Jul 22, 2020 at 8:20
  • You actually answered this one at unix.stackexchange.com/a/397444/5132 a while back, Kusalananda. (-:
    – JdeBP
    Jul 22, 2020 at 9:10
  • There are other issues. The -lr options to ls must be before the filenames. The awk will only show a single line from the pipe, not the first line of each item in the ls command. Jul 22, 2020 at 9:14
  • Another general issue. ssh has a nasty habit of assuming the remote command will read the local stdin, so it reads-ahead (the first 2048 bytes, IIRC) and sends that over with the command. If the ssh is within a script loop (e.g. a list of hosts), the first ssh consumes a chunk of that list. See man ssh, the -f and -n options. Personally, I also explicitly redirect ssh </dev/null within loops too (belt and braces philosophy). Jul 23, 2020 at 10:01

1 Answer 1


The command which is supposed to format the text is within double quotes and is thus treated as text to print. To print the result of a command instead put it inside $() (command substitution), like:

printf "$(printf foo | awk 'NR==1') bar"

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