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

security bash stdin
user283145 picture user283145 · Jan 23, 2011 · Viewed 86.3k times · Source

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 I pass a variable to the stdin of a command? Or, if it's not possible, how to correctly use temporary files for such task?

Answer

Martin picture Martin · Jan 23, 2011

Passing a value to stdin 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.

Related questions

What is the best practice for dealing with passwords in git repositories?

I've got a little Bash script that I use to access twitter and pop up a Growl notification in certain situations. What's the best way to handle storing my password with the script? I would like to commit this script …

Read More
How to use Google application-specific password in script?

Since enabling 2-factor authentication (aka. 2-step verification) on Google, my Google export scripts no longer work. The computer is verified and trusted, but somehow the scripts are not. In effect, every time the cron job is run I receive a …

Read More
How can I prevent SQL injection in PHP?

If user input is inserted without modification into an SQL query, then the application becomes vulnerable to SQL injection, like in the following example: $unsafe_variable = $_POST['user_input']; mysql_query("INSERT INTO `table` (`column`) VALUES ('$unsafe_variable')"); That's …

Read More