BASH - Function to get user password - bash

Question

I created the following BASH script that works perfectly for getting a password from the user:
while [ -z "$PASSWORD" ]
do
echo "Please enter a password:"
read -s PASSWORD1
echo "Please re-enter the password to confirm:"
read -s PASSWORD2
if [ "$PASSWORD1" = "$PASSWORD2" ]; then
PASSWORD=$PASSWORD1
else
# Output error message in red
red='\033[0;31m'
NC='\033[0m' # No Color
echo ""
echo -e "${red}Passwords did not match!${NC}"
fi
done
# This is just here to prove script works
echo "password is: $PASSWORD"
However, if I place it in a function, it stops working:
function getPasswordFromUser()
{
while [ -z "$PASSWORD" ]
do
echo "Please enter a password:"
read -s PASSWORD1
echo "Please re-enter the password to confirm:"
read -s PASSWORD2
if [ "$PASSWORD1" = "$PASSWORD2" ]; then
PASSWORD=$PASSWORD1
else
# Output error message in red
red='\033[0;31m'
NC='\033[0m' # No Color
echo ""
echo -e "${red}Passwords did not match!${NC}"
fi
done
echo $PASSWORD
}
PASSWORD=$(getPasswordFromUser)
# This is just here to check if script worked
echo "got password $PASSWORD"
If I change the call to the function from PASSWORD=$(getPasswordFromUser) to: getPasswordFromUser; then the method starts "working" but the password is output to the screen, and I haven't captured it.
Is there a way to update this BASH script so that I can call a function to get a password from the user without the password ever being displayed in the terminal?
In case it matters, this is for Debian/Ubuntu.

Answer 1

If you call the function like myvar=$(myfunction) it will catch the first echo statement.
What you can do, instead, is to define a variable within the function and then access it. There is no scope in bash, so you will be able to access it once the function has been executed.
See an example on each one of them:
$ cat a
#!/bin/bash
function myf()
{
echo "heeeiiii"
echo "hellO"
}
function myf2()
{
echo "lets define var MYTEST"
MYTEST="this is my test"
}
r=$(myf)
echo "this is myf: $r"
echo "MYTEST=$MYTEST"
myf2
echo "MYTEST=$MYTEST"
Execution:
$ ./a
this is myf: heeeiiii
hellO
MYTEST=
lets define var MYTEST
MYTEST=this is my test

Answer 2

Most of the output in your function should be written to standard error, not standard output.
getPasswordFromUser()
{
while [ -z "$PASSWORD" ]
do
echo "Please enter a password:" >&2
read -s PASSWORD1
echo "Please re-enter the password to confirm:" >&2
read -s PASSWORD2
if [ "$PASSWORD1" = "$PASSWORD2" ]; then
PASSWORD=$PASSWORD1
else
# Output error message in red
red='\033[0;31m'
NC='\033[0m' # No Color
echo -e "\n${red}Passwords did not match!${NC}" >&2
fi
done
echo "$PASSWORD"
}
Also, be sure to quote $PASSWORD in the final line; someone may use multiple runs of whitespace or shell glob characters in their password!

Answer 3

Your first attempt is correct:
~$ PASSWORD=$(getPasswordFromUser)
~$ echo $PASSWORD
Please enter a password: Please re-enter the password to confirm: a
You just don't see the "Please enter a password" because it is captured by the $(..).
You have several possibilities:
use a global variable
use different descriptor
have the function take a variable as the first arg and modify the variable with the string you want to return.
Some examples can be found in how to return a string value from a bash function, the Advanced Bash-Scripting Guide, or the linux journal.