How to pass the value of a variable to the stdin of a command?

Question

I\'m writing a shell script that should be somewhat secure i.e. does not pass secure data through parameters of commands and preferably does not use temporary files. How can

Answer 1

As per Martin's answer, there is a bash feature called Here Strings (which itself is a variant of the more widely supported Here Documents feature).

http://www.gnu.org/software/bash/manual/bashref.html#Here-Strings

3.6.7 Here Strings

A variant of here documents, the format is:

<<< word

The word is expanded and supplied to the command on its standard input.

Note that Here Strings would appear to be bash-only, so, for improved portability, you'd probably be better off with the original Here Documents feature, as per PoltoS's answer:

( cat <<EOF
$variable
EOF
) | cmd

Or, a simpler variant of the above:

(cmd <<EOF
$variable
EOF
)

You can omit ( and ), unless you want to have this redirected further into other commands.

Answer 2

Try this:

echo "$variable" | command

Answer 3

(cat <<END
$passwd
END
) | command

The cat is not really needed, but it helps to structure the code better and allows you to use more commands in parentheses as input to your command.

Answer 4

This robust and portable way has already appeared in comments. It should be a standalone answer.

printf '%s' "$var" | my_cmd

or

printf '%s\n' "$var" | my_cmd

Notes:

• It's better than echo, reasons are here: Why is printf better than echo?
• printf "$var" is wrong. The first argument is format where various sequences like %s or \n are interpreted. To pass the variable right, it must not be interpreted as format.

• Usually variables don't contain trailing newlines. The former command (with %s) passes the variable as it is. However tools that work with text may ignore or complain about an incomplete line (see Why should text files end with a newline?). So you may want the latter command (with %s\n) which appends a newline character to the content of the variable. Non-obvious facts:

  • Here string in Bash (<<<"$var" my_cmd) does append a newline.
  • Any method that appends a newline results in non-empty stdin of my_cmd, even if the variable is empty or undefined.

Answer 5

I liked Martin's answer, but it has some problems depending on what is in the variable. This

your-command <<< """$your_variable"""

is better if you variable contains " or !

Answer 6

Just do:

printf "$my_var" | my_cmd

If the var doesn't contain spaces then the quotes may be omitted.
If using bash then you may also do:

echo -n "$my_var" | my_cmd

Avoid using echo without -n because it will pipe the vraiable with an added linebreak at the end.