How do you find the original user through multiple

问题:

When running a script via sudo or su I want to get the original user. This should happen regardless of multiple sudo or su runs inside of each other and specifically sudo su -.

回答1:

Results:

Use who am i | awk '{print $1}' OR logname as no other methods are guaranteed.

Logged in as self:

evan> echo $USER
evan
evan> echo $SUDO_USER

evan> echo $LOGNAME
evan
evan> whoami
evan
evan> who am i | awk '{print $1}'
evan
evan> logname
evan
evan>

Normal sudo:

evan> sudo -s
root> echo $USER
root
root> echo $SUDO_USER
evan
root> echo $LOGNAME
root
root> whoami
root
root> who am i | awk '{print $1}'
evan
root> logname
evan
root>

sudo su - :

evan> sudo su -
[root ]# echo $USER
root
[root ]# echo $SUDO_USER

[root ]# echo $LOGNAME
root
[root ]# whoami
root
[root ]# who am i | awk '{print $1}'
evan
[root ]# logname
evan
[root ]#

sudo su -; su tom :

evan> sudo su -
[root ]# su tom
tom$ echo $USER
tom
tom$ echo $SUDO_USER

tom$ echo $LOGNAME
tom
tom$ whoami
tom
tom$ who am i | awk '{print $1}'
evan
tom$ logname
evan
tom$

回答2:

There's no perfect answer. When you change user IDs, the original user ID is not usually preserved, so the information is lost. Some programs, such as logname and who -m implement a hack where they check to see which terminal is connected to stdin, and then check to see what user is logged in on that terminal.

This solution often works, but isn't foolproof, and certainly shouldn't be considered secure. For example, imagine if who outputs the following:

tom     pts/0        2011-07-03 19:18 (1.2.3.4)
joe     pts/1        2011-07-03 19:10 (5.6.7.8)

tom used su to get to root, and runs your program. If STDIN is not redirected, then a program like logname will output tom. If it IS redirected (e.g. from a file) as so:

logname < /some/file

Then the result is "no login name", since the input isn't the terminal. More interestingly still, though, is the fact that the user could pose as a different logged in user. Since Joe is logged in on pts/1, Tom could pretend to be him by running

logname < /dev/pts1

Now, it says joe even though tom is the one who ran the command. In other words, if you use this mechanism in any sort of security role, you're crazy.

回答3:

This is a ksh function I wrote on HP-UX. I don't know how it will work with Bash in Linux. The idea is that the sudo process is running as the original user and the child processes are the target user. By cycling back through parent processes, we can find the user of the original process.

#
# The options of ps require UNIX_STD=2003.  I am setting it
# in a subshell to avoid having it pollute the parent's namespace.
#
function findUser
{
    thisPID=$$
    origUser=$(whoami)
    thisUser=$origUser
    while [ "$thisUser" = "$origUser" ]
    do
        ( export UNIX_STD=2003; ps -p$thisPID -ouser,ppid,pid,comm ) | grep $thisPID | read thisUser myPPid myPid myComm
        thisPID=$myPPid
    done
    if [ "$thisUser" = "root" ]
    then
        thisUser=$origUser
    fi
    if [ "$#" -gt "0" ]
    then
        echo $origUser--$thisUser--$myComm
    else
        echo $thisUser
    fi
    return 0
}

I know the original question was from a long time ago but people (such as me) are still asking and this looked like a good place to put the solution.

回答4:

How about using logname(1) to get the user's login name?

回答5:

THIS_USER=`pstree -lu -s $$ | grep --max-count=1 -o '([^)]*)' | head -n 1 | sed 's/[()]//g'`

That's the only thing that worked for me.

回答6:

user1683793's findUser() function ported to bash and extended so it returns usernames stored in NSS libraries as well.

#!/bin/bash

function findUser() {
    thisPID=$$
    origUser=$(whoami)
    thisUser=$origUser

    while [ "$thisUser" = "$origUser" ]
    do
        ARR=($(ps h -p$thisPID -ouser,ppid;))
        thisUser="${ARR[0]}"
        myPPid="${ARR[1]}"
        thisPID=$myPPid
    done

    getent passwd "$thisUser" | cut -d: -f1
}

user=$(findUser)
echo "logged in: $user"

回答7:

cycling back and giving a list of users

based on user1683793's answer

By exlcuding non-TTY processes, I skip root as the initiator of the login. I'm not sure if that may exlcude too much in some case

#!/bin/ksh
function findUserList
{
    typeset userList prevUser thisPID thisUser myPPid myPid myTTY myComm
    thisPID=$$                 # starting with this process-ID
    while [ "$thisPID" != 1 ]  # and cycling back to the origin
    do
        (  ps -p$thisPID -ouser,ppid,pid,tty,comm ) | grep $thisPID | read thisUser myPPid myPid myTTY myComm
        thisPID=$myPPid
        [[ $myComm =~ ^su ]] && continue        # su is always run by root -> skip it
        [[ $myTTY == '?' ]] && continue         # skip what is running somewhere in the background (without a terminal)
        if [[ $prevUser != $thisUser ]]; then   # we only want the change of user
                prevUser="$thisUser"            # keep the user for comparing
                userList="${userList:+$userList }$thisUser"  # and add the new user to the list
        fi
        #print "$thisPID=$thisUser: $userList -> $thisUser -> $myComm " >&2
    done
    print "$userList"
    return 0
}

logname or who am i didn't give me the desired answer, especially not in longer lists of su user1, su user2, su user3, ...

I know the original question was from a long time ago but people (such as me) are still asking and this looked like a good place to put the solution.

回答8:

Alternative to calling ps multiple times: do one pstree call

pstree -lu -s $$ | grep --max-count=1 -o '([^)]*)' | head -n 1

output (when logged in as even): (evan)

pstree arguments:

• -l: long lines (not shortening)
• -u: show when user changes as (userName)
• -s $$: show parents of this process

Get the first user change (which is login) with grep -o and head.

limitation:the command may not contain any braces () (it does not normally)