Hiding user input on terminal in Linux script - linux

Question

I have bash script like the following:
#!/bin/bash
echo "Please enter your username";
read username;
echo "Please enter your password";
read password;
I want that when the user types the password on the terminal, it should not be displayed (or something like *******) should be displayed). How do I achieve this?

Answer 1

Just supply -s to your read call like so:
$ read -s PASSWORD
$ echo $PASSWORD

Answer 2

Update
In case you want to get fancy by outputting an * for each character they type, you can do something like this (using andreas' read -s solution):
unset password;
while IFS= read -r -s -n1 pass; do
if [[ -z $pass ]]; then
echo
break
else
echo -n '*'
password+=$pass
fi
done
Without being fancy
echo "Please enter your username";
read username;
echo "Please enter your password";
stty -echo
read password;
stty echo

Answer 3

you can use stty to disable echo
this solution works without bash or certain features from read
stty_orig=$(stty -g)
stty -echo
read password
stty $stty_orig
If you use this in a shell script then also set an exit handler which restores echo:
#! /bin/sh
stty_orig=$(stty -g)
trap "stty ${stty_orig}" EXIT
stty -echo
...
this is to make sure echo is restored regardless of how the script exits. otherwise the echo will stay off if the script errors out.
to turn echo back on manually type the following command
stty echo
you will have to type blindly because you do not see what you type.
i suggest to press ctrl+c first to clear anything else you might have typed before.
trivia
echo means to echo your typed input back to your screen.
this is from the time we worked on teletypewriters (that is what the tty means). a teletypewriter is like a typewriter but connected to another teletypewriter or computer. typically via telephone cable.
the workflow on a teletypewriter is roughly as follows: you type in your command (or message for the other side). then the teletypewriter will print the response from the other side.
when you work on a teletypewriter you see your input as you type. this is because the teletypewriter is also a typewriter and as such prints the characters as you press them.
when teletypewriters where replaced by screens there was no longer a typewriter which types your input. instead we had to deliberate code an "echo" function which prints your input as you type.
i do not know whether stty -echo also disabled printing on a teletypewriter.
see here for a teletypewriter in action: https://www.youtube.com/watch?v=2XLZ4Z8LpEE (first part is restoration. action starting at about 12 minutes in)
more teletypewriter restoration: https://www.youtube.com/playlist?list=PL-_93BVApb5-9eQLTCk9xx16RAGEYHH1q

Answer 4

Here's a variation on #SiegeX's excellent *-printing solution for bash with support for backspace added; this allows the user to correct their entry with the backspace key (delete key on a Mac), as is typically supported by password prompts:
#!/usr/bin/env bash
password=''
while IFS= read -r -s -n1 char; do
[[ -z $char ]] && { printf '\n' >/dev/tty; break; } # ENTER pressed; output \n and break.
if [[ $char == $'\x7f' ]]; then # backspace was pressed
# Remove last char from output variable.
[[ -n $password ]] && password=${password%?}
# Erase '*' to the left.
printf '\b \b' >/dev/tty
else
# Add typed char to output variable.
password+=$char
# Print '*' in its stead.
printf '*' >/dev/tty
fi
done
Note:
As for why pressing backspace records character code 0x7f: "In modern systems, the backspace key is often mapped to the delete character (0x7f in ASCII or Unicode)" https://en.wikipedia.org/wiki/Backspace
\b \b is needed to give the appearance of deleting the character to the left; just using \b moves the cursor to the left, but leaves the character intact (nondestructive backspace). By printing a space and moving back again, the character appears to have been erased (thanks, The "backspace" escape character '\b': unexpected behavior?).
In a POSIX-only shell (e.g., sh on Debian and Ubuntu, where sh is dash), use the stty -echo approach (which is suboptimal, because it prints nothing), because the read builtin will not support the -s and -n options.

Answer 5

A bit different from (but mostly like) #lesmana's answer
stty -echo
read password
stty echo
simply: hide echo
do your stuff
show echo

Answer 6

I always like to use Ansi escape characters:
echo -e "Enter your password: \x1B[8m"
echo -e "\x1B[0m"
8m makes text invisible and 0m resets text to "normal." The -e makes Ansi escapes possible.
The only caveat is that you can still copy and paste the text that is there, so you probably shouldn't use this if you really want security.
It just lets people not look at your passwords when you type them in. Just don't leave your computer on afterwards. :)
NOTE:
The above is platform independent as long as it supports Ansi escape sequences.
However, for another Unix solution, you could simply tell read to not echo the characters...
printf "password: "
let pass $(read -s)
printf "\nhey everyone, the password the user just entered is $pass\n"

Answer 7

Here is a variation of #SiegeX's answer which works with traditional Bourne shell (which has no support for += assignments).
password=''
while IFS= read -r -s -n1 pass; do
if [ -z "$pass" ]; then
echo
break
else
printf '*'
password="$password$pass"
fi
done

Answer 8

Get Username and password
Make it more clear to read but put it on a better position over the screen
#!/bin/bash
clear
echo
echo
echo
counter=0
unset username
prompt=" Enter Username:"
while IFS= read -p "$prompt" -r -s -n 1 char
do
if [[ $char == $'\0' ]]; then
break
elif [ $char == $'\x08' ] && [ $counter -gt 0 ]; then
prompt=$'\b \b'
username="${username%?}"
counter=$((counter-1))
elif [ $char == $'\x08' ] && [ $counter -lt 1 ]; then
prompt=''
continue
else
counter=$((counter+1))
prompt="$char"
username+="$char"
fi
done
echo
unset password
prompt=" Enter Password:"
while IFS= read -p "$prompt" -r -s -n 1 char
do
if [[ $char == $'\0' ]]; then
break
elif [ $char == $'\x08' ] && [ $counter -gt 0 ]; then
prompt=$'\b \b'
password="${password%?}"
counter=$((counter-1))
elif [ $char == $'\x08' ] && [ $counter -lt 1 ]; then
echo
prompt=" Enter Password:"
continue
else
counter=$((counter+1))
prompt='*'
password+="$char"
fi
done

Answer 9

A variation on both #SiegeX and #mklement0's excellent contributions: mask user input; handle backspacing; but only backspace for the length of what the user has input (so we're not wiping out other characters on the same line) and handle control characters, etc... This solution was found here after so much digging!
#!/bin/bash
#
# Read and echo a password, echoing responsive 'stars' for input characters
# Also handles: backspaces, deleted and ^U (kill-line) control-chars
#
unset PWORD
PWORD=
echo -n 'password: ' 1>&2
while true; do
IFS= read -r -N1 -s char
# Note a NULL will return a empty string
# Convert users key press to hexadecimal character code
code=$(printf '%02x' "'$char") # EOL (empty char) -> 00
case "$code" in
''|0a|0d) break ;; # Exit EOF, Linefeed or Return
08|7f) # backspace or delete
if [ -n "$PWORD" ]; then
PWORD="$( echo "$PWORD" | sed 's/.$//' )"
echo -n $'\b \b' 1>&2
fi
;;
15) # ^U or kill line
echo -n "$PWORD" | sed 's/./\cH \cH/g' >&2
PWORD=''
;;
[01]?) ;; # Ignore ALL other control characters
*) PWORD="$PWORD$char"
echo -n '*' 1>&2
;;
esac
done
echo
echo $PWORD