zsh script not emitting expected enable/disable output-CodePudding

I've got this following bash script, in witch for every user loggend in a mac i will check if the secure token status is set to enabled. Then I will count, with grep -ic command, how many time the word enable is present. Then, if the word enable is not equal to 0 I will print the username and the word enabled otherwise disable. This is the output i've got:

luigiMac
martaMac
provamacmarco:Disabled

This is the script:

#!/usr/bin/env zsh

result=""
enabled="Enabled"
disabled="Disabled"
users=$(dscl . -list /Users | grep -v -e '_' -e root -e nobody -e daemon)

for item_ in "${users[@]}"; do
    newUsers =("$item_")
    #echo $newUsers
done

for i in "${newUsers[@]}";do
#echo $i[4]
tokenStatus =$(sysadminctl -adminUser "" -adminPassword "" -secureTokenStatus "$i" 2>&1 | grep -ic "enabled")
print $tokenStatus
if [[ "${tokenStatus[@]}" -ne 0 ]]
then
result="${i}:${enabled}"
echo "${result}"
else
result="${i}:${disabled}"
echo "${result}"
fi
done

#echo "${result}"

I don't understand why the script doesn't print enable or disable for the other username.

CodePudding user response：

I'm no zsh expert but IMO you're doing a lot of unnecessary things in your script; you could somewhat simplify it to:

dscl . -list /Users |
grep -vE -e '^_' -e '^(root|daemon|nobody)$' |
while IFS=$'\n' read -r user
do
    if sysadminctl ... -secureTokenStatus "$user" 2>&1 |
       grep -q 'ENABLED'
    then
        result=Enabled
    else
        result=Disabled
    fi
    printf '%s:%s\n' "$user" "$result"
done