Bash printf value does not show up or is cut off-CodePudding

I trying to get a value from a command into a var
and then print it our using printf.

Problem: i got the value in the var but with printf it does not appear
or is cut off.

INFO: In my script im calling redis-cli info memory
and to check whats wrong i tried a call on vmstat -s.

Working vmstat test:

format="%-16s | %-16s"
container_name="some_name"
used_memory=$(vmstat -s | sed -n "s/^\(.*\) K used memory.*$/\1/p")
row=$(printf "${format}" "${container_name}" "${used_memory}")
echo "${row}"

Output: some_name | 11841548

The actual script that is not working:

format="%-50s | %-16s"
container_name="totally_secret_container_name_1"
used_memory=$(docker exec -it "${container_name}" redis-cli info memory | sed -n "s/^used_memory_human:\(.*\)$/\1/p")
row=$(printf "${format}" "${container_name}" "${used_memory}")
echo "${row}"

Output: ecret_container_name_1 | 1.08M

Weird is than when i set the format to format="%-50s | %-1s"
then it works - the container name (left value) gets printed correctly.

What happen here?
How can i fix this?

Thanks for your time!

CodePudding user response：

You need to remove the \r characters in the output that are causing it to go back to the beginning of the line and overwrite.

used_memory=$(docker exec -it "${container_name}" redis-cli info memory | sed -n "s/^used_memory_human:\(.*\)$/\1/p")
used_memory=${used_memory//$'\r'/}
row=$(printf "${format}" "${container_name}" "${used_memory}")

This uses the bash ${variable//old/new} replacement operator.