I have a script which contains the code block:
cat << EOF > new_script.sh ... echo "$(pwd)" >> log.txt ... EOF
The script new_script.sh is set to run at a later time. Bash recognizes the $(pwd) within the script and evaluates it before it looks at the entire EOF block, so the pwd of the current directory is output instead of the pwd of new_script.sh when it is run. Why is this the case (what logic does bash use to know to evaluate $(command)) and what is the best solution to this?
By adding an escape $,
\$ , you can solve this issue.
cat << EOF > new_script.sh ... echo "\$(pwd)" >> log.txt ... EOF