How can pass the value of a variable to the standard input of a command?

Question

Passing a value to standard input in Bash is as simple as:

your-command <<< "$your_variable"

Always make sure you put quotes around variable expressions!

Be cautious, that this will probably work only in bash and will not work in sh.

Simple, but error-prone: using echo

Something as simple as this will do the trick:

echo "$blah" | my_cmd

Do note that this may not work correctly if $blah contains -n, -e, -E etc; or if it contains backslashes (bash's copy of echo preserves literal backslashes in absence of -e by default, but will treat them as escape sequences and replace them with corresponding characters even without -e if optional XSI extensions are enabled).

More sophisticated approach: using printf

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

This does not have the disadvantages listed above: all possible C strings (strings not containing NULs) are printed unchanged.

Note that the 'echo "$var" | command operations mean that standard input is limited to the line(s) echoed. If you also want the terminal to be connected, then you'll need to be fancier:

{ echo "$var"; cat - ; } | command

( echo "$var"; cat -   ) | command

This means that the first line(s) will be the contents of $var but the rest will come from cat reading its standard input. If the command does not do anything too fancy (try to turn on command line editing, or run like vim does) then it will be fine. Otherwise, you need to get really fancy - I think expect or one of its derivatives is likely to be appropriate.

The command line notations are practically identical - but the second semi-colon is necessary with the braces whereas it is not with parentheses.

(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.

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.

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 !.