Bash cannot access a variable inside function when running as another user-CodePudding

I have the below script and I am trying to access logtime variable when calling the function as userA but the $logtime inside the function is not getting the value. This works when I am not switching the user.

logtime=`date  "%Y-%m-%d_%H%M%S"`

runcommand()
{
echo " Log time is $logtime"
}

export -f runcommand
su userA -c "bash -c runcommand >>  compile-$logtime.txt"

$logtime is not getting assigned inside the function when called as userA

cat compile-2022-11-07_121225.txt
Log time is

CodePudding user response：

declare -f funcname emits source that, when evaluated, recreates your function.

declare -p varname emits source that, when evaluated, recreates your variable.

These both can be added into the content you run on the other side of a privilege boundary. Thus:

sudo -u userA bash <<EOF
$(declare -f runcommand)
$(declare -p logtime)
runcommand >>"compile-$logtime.txt" # note it's the outer shell expanding $logtime
EOF

...or, without the heredoc (the switch from double to single quotes lets the inner shell expand $logtime here, contrary to the above):

sudo -u userA bash -c "$(declare -f runcommand); $(declare -p logtime)"'; runcommand >>compile-$logtime.txt'

If you do stick with su, though, the easiest fix is just to export logtime.

export logtime
export -f runcommand
su userA -c 'bash -c runcommand >>"compile-$logtime.txt"'

This is generally silly, though -- you have some other shell invoking bash, so the real invocation looks like sh -c 'bash -c runcommand'. Taking out the shell in the middle makes life easier for everyone.

CodePudding user response：

Variables will remain inside of the running scritp, even accessing them from the current user, won't be possible.

you may try to first source the script file, then execute it