An escaping of the variable within bash script

Question

My bash script writes an another bash script using printf.

printf \"#!/bin/bash
HOME=${server}
file=gromacs*
file_name=\\$(basename \"\\${file}\")
date=\\$(date          


        

Answer 1

You have to escapes in printf with esscaps, e. g.:

date=\$(date +"%%m_%%d_%%Y")

should print date=\$(date +"%m_%d_%Y"). But you should avoid using printf, because you don't use it's capabilities. Instead you could cat the string to the file:

cat > ${output}/collecter.sh <<'END'
HOME=${server}
...
done
END

This would allow you to avoid many escapes, and make the code more readable.