Using echo in Bash with variable

Question

I want to use the line below to change a password. When I put actual values instead of variables, the command works perfectly.

echo -e "$sshpass\n$sshpass | (passwd --stdin root)"

$sshpass is a variable containing a password.

I tried the following to make commands work with no luck:

echo -e "/$sshpass\n/$sshpass | (passwd --stdin root)"
echo -e "$sshpass\n$sshpass | (passwd --stdin root)"
echo -e "'$sshpass'\n/'$sshpass' | (passwd --stdin root)"

How can I make it work?

Are you sure the first one works when you use echo -e "xyz\nxyz | (passwd --stdin root)"? It shouldn't change the password. — Jonathan Leffler
– Jonathan Leffler, Commented Aug 18, 2013 at 15:25

choroba · Accepted Answer · 2013-08-18 15:21:38Z

5

Do not include the rest of the pipeline in the double quotes. The subshell (parentheses) is not needed.

echo -e "$sshpass\n$sshpass" | passwd --stdin root

answered Aug 18, 2013 at 15:21

choroba

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Comments

Peter Mortensen · Accepted Answer · 2014-05-25 21:51:38Z

1

Use this:

echo -e "$sshpass\n$sshpass" | passwd --stdin root

edited May 25, 2014 at 21:51

Peter Mortensen

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answered Aug 18, 2013 at 15:22

Jonathan Leffler

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