How to run through password function twice with different variables? - bash

Question

I have a password function that I borrowed from How do I echo stars (*) when reading password with read?
I tried to adapt it so that I can run through the function twice to do a password confirmation and then evaluate the 2 passwords to determine if they match but I seem to be missing some basics of how bash works in this case.
I tried replacing PASSWORD with $1 but kept getting command not found errors
passWord() {
unset PASSWORD
unset CHARCOUNT
stty -echo
CHARCOUNT=0
while IFS= read -p "$PROMPT" -r -s -n 1 CHAR; do
# Enter - accept password
if [[ $CHAR == $'\0' ]] ; then
break
fi
# Backspace
if [[ $CHAR == $'\177' ]] ; then
if [ $CHARCOUNT -gt 0 ] ; then
CHARCOUNT=$((CHARCOUNT-1))
PROMPT=$'\b \b'
PASSWORD="${PASSWORD%?}"
else
PROMPT=''
fi
else
CHARCOUNT=$((CHARCOUNT+1))
PROMPT='*'
PASSWORD+="$CHAR"
fi
done
stty echo; echo
${1}=${PASSWORD}
}
echo -n "Enter the password > "
passWord passOne
echo -n "Please re-enter the password > "
passWord passTwo
if [[ $passOne == $passTwo ]]; then
PASSWORD=$passOne
else
echo "Passwords did not match, please try again."
fi
Update
Here is the script with the latest updates
#!/bin/bash
passWord() {
unset password
local prompt char
stty -echo
charcount=0
while IFS= read -p "$prompt" -r -s -n 1 CHAR; do
# Enter - accept password
if [[ $char == $'\0' ]] ; then
break
fi
# Backspace
if [[ $char == $'\177' ]] ; then
if [ $charcount -gt 0 ] ; then
charcount=$((CHARCOUNT-1))
prompt=$'\b \b'
password="${password%?}"
else
prompt=''
fi
else
charcount=$((charcount+1))
prompt='*'
password+="$char"
fi
done
stty echo; echo
}
echo -n "Enter the password > "
passWord
pass1=$password
echo -n "Please re-enter the password > "
passWord
pass2=$password
if [[ "$pass1" == "$pass2" ]]; then
PassWord=$pass1
else
echo "Passwords did not match, please try again."
fi

Answer 1

You are missing a declaration of your shell.
Please add a shebang as the first line:
#!/bin/bash
The assignment of variables (the line ${1}=${PASSWORD}) doesn't work.
One way to solve it (not recomended) is to add eval:
eval "${1}=${PASSWORD}" # don't use quite risky.
But as that makes any input a security issue, you should use some other line.
One solution is to use declare (bash 4.2+):
declare -g "${1}=${PASSWORD}"
The -g is required (required and available since bash 4.2) to change General variables (not local to the function).
Or use printf (since bash 3.1):
printf -v "${1}" '%s' "${PASSWORD}"
Other than that, you should add a local command for variables used inside the function to avoid conflicts with external variables and should add a PROMPT='' just before the loop to avoid the printing of an initial asterisk when calling the function a second time.
It should be said that using variables in CAPS should be avoided. Variables in CAPS denote environment variables, the rest of variables use lower case to avoid conflicts.