Why is my awk print not showing up on the terminal - bash

Question

I have the following script which does a "which -a" on a command then a "ls -l" to let me know if it's a link or not .. ie "grep" since I have gnu commands installed (Mac with iTerm).
#!/usr/bin/env bash
which -a $1 | xargs -I{} ls -l "{}" \
| awk '{for (i = 1; i < 9; i++) $i = ""; sub(/^ */, ""); print}'
When I run this from the script "test grep" I receive no output, but when I run it via "bash -x test grep" I receive the following:
bash -x test grep
+ which -a grep
+ xargs '-I{}' ls -l '{}'
+ awk '{for (i = 1; i < 9; i++) $i = ""; sub(/^ */, ""); print}'
/usr/local/bin/grep -> ../Cellar/grep/3.1/bin/grep
/usr/bin/grep
The last 2 lines is what I'm looking to display. Thought this would be easier to do ;-) .. I also tried appending the pipe thinking printf would fix the issue:
| while read path
do
printf "%s\n" "$path"
done
Thanks and .. Is there a better way to get what I need?

Answer 1

The problem is that you named your script test.
If you want to run a command that's not in your PATH, you need to specify the directory it's in, e.g. ./test.
You're not getting an error for trying to run test because there is a built-in bash command called test that is used instead. For extra confusion, the standard test produces no output.
In conclusion:
Use ./ to run scripts in the current directory.
Never call your test programs test.

Answer 2

Thanks for the never naming a script "test" .. old habits are hard to break (I came from a non-unix background.
I ended with the following
for i in $(which -a $1)
do
stat $i | awk NR==1{'$1 = ""; sub(/^ */, ""); print}'
done
or simpler
for i in $(which -a $1)
do
stat -c %N "$i"
done

Answer 3

Consider the following shell function:
cmdsrc() {
local cmd_file cmd_file_realpath
case $(type -t -- "$1") in
file) cmd_file=$(type -P -- "$1")
if [[ -L "$cmd_file" ]]; then
echo "$cmd_file is a symlink" >&2
elif [[ -f "$cmd_file" ]]; then
echo "$cmd_file is a regular file" >&2
else
echo "$cmd_file is not a symlink or a regular file" >&2
fi
cmd_file_realpath=$(readlink -- "$cmd_file") || return
if [[ $cmd_file_realpath != "$cmd_file" ]]; then
echo "...the real location of the executable is $cmd_file_realpath" >&2
fi
;;
*) echo "$1 is not a file at all: $(type -- "$1")" >&2 ;;
esac
}
...used as such:
$ cmdsrc apt
/usr/bin/apt is a symlink
...the real location of the executable is /System/Library/Frameworks/JavaVM.framework/Versions/A/Commands/apt
$ cmdsrc ls
/bin/ls is a regular file
$ cmdsrc alias
alias is not a file at all: alias is a shell builtin

Answer 4

Took some suggestions and came up with the following:
The prt-underline is just a fancy printf function. I decided not to go with readline since the ultimate command resolution may be unfamiliar to me and I only deal with regular files .. so does't handle every situation but in the end gives me the output I was looking for. Thanks for all the help.
llt ()
{
case $(type -t -- "$1") in
function)
prt-underline "Function";
declare -f "$1"
;;
alias)
prt-underline "Alias";
alias "$1" | awk '{sub(/^alias /, ""); print}'
;;
keyword)
prt-underline "Reserved Keyword"
;;
builtin)
prt-underline "Builtin Command"
;;
*)
;;
esac;
which "$1" &> /dev/null;
if [[ $? = 0 ]]; then
prt-underline "File";
for i in $(which -a $1);
do
stat "$i" | awk 'NR==1{sub(/^ File: /, ""); print}';
done;
fi
}