passing command as argument in bash

Question

I have a simple script that takes the first argument from the command line and prints hello, first_arg.

But when I pass |ls, it is printing files in the current directory.

My script file

#! /bin/bash

echo "Hello, $1."

I tried multiple things

echo "Hello, '$'"
echo "Hello," "$1"
printf "Hello %s" "$1"

I want output like Hello, |ls

I checked quoting and this answer https://unix.stackexchange.com/questions/171346/security-implications-of-forgetting-to-quote-a-variable-in-bash-posix-shells

Edit

but It is not happening with input like

echo "input: "
read name
echo "Hello, $name."  # output Hello, |ls
echo "Hello, " $name  # output Hello, |ls

Simply quote "|ls" on the command line to prevent expansion (piping to ls) by the shell. E.g. bash yourscript "|ls" (single or double quotes are fine here) — David C. Rankin
– David C. Rankin, Commented Nov 7, 2021 at 3:20
If you invoke the script with quotes around the argument, as in script '|ls', it will work. If you don't use single or double quotes, then the command line is read as 'output of your script piped to ls', and ls doesn't read its standard input. So the problem is not inside the script — it is all in how you invoke it. — Jonathan Leffler
– Jonathan Leffler, Commented Nov 7, 2021 at 3:21
No. The expansion is preformed by the shell when your script is invoked. Meaning the shell would have already set up to pipe the result of your script to ls before the script was even started. By quoting the argument you tell the shell to treat it as a string (and argument to your script) and not further shell expressions. — David C. Rankin
– David C. Rankin, Commented Nov 7, 2021 at 3:52

score 1 · Accepted Answer · 2021-11-07 05:52:36Z

1

You need to quote the pipe when invoking the script. Otherwise the shell will arrange for the standard output your script to be sent standard input of ls and you will end up with the output as if you hand run ls on it's own. Either of these would do:

./simple.sh \|ls
./simple.sh '|ls'
./simple.sh "|ls"

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