I would like to programmatically answer a password prompt using only shell script when prompted. Is it possible to automate this task in this way? I want to avoid other tools entirely - so no expect or similar - I would like to achieve this simply using shell script.


1 Answer 1


Since you do not specify which shell, I'll just cheat. You can find the following and much else on grml.org's zsh-lover's command reference man page.

  $ zmodload zsh/zpty
  $ zpty PW passwd $1
  $ zpty PW passwd $1
# ``-r'': read the output of the command name.
# ``z'' : Parameter
  $ zpty -r PW z '*password:'
# send the to command name the given strings as input
  $ zpty -w PW $2
  $ zpty -r PW z '*password:'
  $ zpty -w PW $2
# The second form, with the -d option, is used to delete commands
# previously started, by supplying a list of their names. If no names
# are given, all commands are deleted. Deleting a command causes the HUP
# signal to be sent to the corresponding process.
  $ zpty -d PW

I even managed to do it myself. Here's a copy+paste from my terminal:

mikeserv@localhost ~ % :                  
sudo zsh -c '
zmodload zsh/zpty
userdel dummy
useradd -m dummy
zpty dummyadd "passwd dummy"
zpty -r dummyadd nl "*UNIX password:"
zpty -w dummyadd "dummypw^M"
zpty -r dummyadd nl "*UNIX password:"
zpty -w dummyadd "dummypw^M"
zpty -r dummyadd nl "*"
printf %s\\n "$nl"
zpty -d dummyadd
userdel: user 'dummy' does not exist
useradd: warning: the home directory already exists.
Not copying any file from skel directory into it.

mikeserv@localhost ~ % ssh dummy@localhost
dummy@localhost's password: 
dummy@localhost ~ % 

It took me a few tries - that's why dummy's home directory was already there - but you can see that when I ran that command that user did not exist, and I scripted a session for setting dummy's password. It's pretty clumsy script granted, but, when it was over, I was able to login as dummy.

So I guess it's possible. Maybe not advisable, but maybe that depends on what you're about.

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