Bash script - "err: command not found"?

Question

I am writing a bash script and I am using a for cycle to check my arguments.

for var in "$@"
do
 test_arg "$var"
done

And this is my test_arg function

function test_arg {
 [ -n "1ドル" ] || err "Empty argument"
 [ -f "1ドル" ] || err "Argument '1ドル' is not a file"
 [ -r "1ドル" ] || err "Data file '1ドル' is not readable"
 [ -s "1ドル" ] || err "Data file '1ドル' is empty"
 egrep -v '^-?([0-9]+|[0-9]*\.[0-9]+)$' "1ドル" && { echo "Bad data format in '1ドル'"; exit 1; }
}

However, when any of these conditions are not met, script only writes out "script.sh: line XX: err: command not found". I am not quite sure about the testing, I am a bash begginer.

Thank you very much for your answers

• Charles Duffy: err "Empty argument" -- err is not a built-in command, so unless you define it as a function, it won't be found.
• Charles Duffy: By the way, don't use the function keyword; it serves no purpose but to make your code incompatible with baseline POSIX shells. Just test_arg() { ...} will do.
• Charles Duffy: (...anyhow, this has nothing to do with -s, or with the [ or test builtins).
• SaintHax: If we are going to point out the function keyword (which I agree with Charles about), don't give do it's own line. It serves no purpose but to make your code harder to read. for var in "$@"; do

Answer 1

Your code depends on a function named err. Consider defining it like so:

err() { echo "$*" >&2; exit 1; }

• Jesse_Pinkman: yes - now it works like a charm - thank you very much sir - will accept as an answer when possible