awk in bash with ls and variable

Question

I wanted to print only the name of files of a specific directory: In this way it works:

ls -g  --sort=size -r /bin | awk '{print $8,$9,$10,$11,$12,$13}'

but if I read the path variable it doesn't work:

read PATH
ls -g  --sort=size -r $(PATH) | awk '{print $8,$9,$10,$11,$12,$13}'
Command 'awk' is available in '/usr/bin/awk'

Answer 1

It should be:

ls -g  --sort=size -r ${PATH} | awk '{print $8,$9,$10,$11,$12,$13}'

Notice the curly braces.

With $(..), it'll execute the command/function named PATH and substitute the result, which is not what you want.

Note that PATH is the poor choice for a variable name as it will overwrite the system variable with the same name and make the system commands unavailable (since the original PATH has gone now). I suggest you use some other name such as my_path:

read my_path
ls -g  --sort=size -r ${my_path} | awk '{print $8,$9,$10,$11,$12,$13}'

Answer 2

This is exactly why you should not use UPPER_CASE_VARS. $PATH is a variable used by the shell to find executables on your system. As soon as you over-write it with user input, your script can no longer find anything that does not reside in whatever the input was. In this case, you entered /bin, so your script can find /bin/ls but awk is not there.

The command_not_found_handle (see /etc/bash.bashrc) stepped in to give you a suggestion.