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Bash argument not expanding in script

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# 1  
Old 10-11-2019
Bash argument not expanding in script

I pass an argument to bash as run. The first command in green executes as expected, however the second in blue fails as the $run does not expand. I tried to escape the variable with \ thinking the quotes were making the literal translation and also "${run}" but both did not work to expand the variable run. What am I missing? Thank you Smilie.

bash script <run>


printf -v cmd_q '(cd path/to/dir/%q*/*/out && exec sshpass -f out.txt scp -- *.txt**/folder)\n' "${array[@]}"
sshpass -f file.txt ssh -o strictHostKeyChecking=no -t "$cmd_q"

printf -v cmd_q '(cd path/to/dir/%q*/*/out && exec sshpass -f out.txt scp -- *.txt*"$run"/folder)\n' "${array[@]}"
sshpass -f file.txt ssh -o strictHostKeyChecking=no -t "$cmd_q"

# 2  
Old 10-11-2019
Try export run=$1 to make the variable visible in child processes. I can't see a mistake in the syntax.
This User Gave Thanks to jim mcnamara For This Post:
# 3  
Old 10-13-2019
Shells don't expand variables within single quotes.
These 2 Users Gave Thanks to RudiC For This Post:
# 4  
Old 10-14-2019
Thank you very much, the below worked Smilie

printf -v cmd_q '(cd path/to/dir/%q*/*/out && exec sshpass -f out.txt scp -- *.txt*'"$run"'/folder)\n' "${array[@]}"
sshpass -f file.txt ssh -o strictHostKeyChecking=no -t "$cmd_q"

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