Why print empty path in shell script?

filter_option=" -name "*" "
file_paths=$(find "." -type f $filter_option)
echo find "." -type f $filter_option
echo $file_paths

JavaScript
​x
 
filter_option=" -name "*" "file_paths=$(find "." -type f $filter_option)echo find "." -type f $filter_optionecho $file_paths​

samuel@:~/…/linux$ ./test.sh
find . -type f -name “*”

samuel@:~/…/linux$

How do I fix this to get find . -type f -name "*" ‘s result?

Answer

I suggest to use an array:

filter_option=(-name '*')
file_paths=$(find "." -type f "${filter_option[@]}")

JavaScript
 
filter_option=(-name '*')file_paths=$(find "." -type f "${filter_option[@]}")​