Identify user in a Bash script called by sudo

If开发者_Go百科 I create the script /root/bin/whoami.sh containing:

#!/bin/bash
whoami

and this script is called by a user with a properly configured sudo, it will indicate

root

Is there a fast way to obtain the actual user in a script, or will I have to resort to parameters passing along this username?

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$SUDO_USER doesn't work if you are using sudo su -.
It also requires multiple checks - if $USER == 'root' then get $SUDO_USER.

Instead of the command whoami use who am i. This runs the who command filtered for the current session. It gives you more info than you need. So, do this to get just the user:

who am i | awk '{print $1}'

Alternatively (and simpler) you can use logname. It does the same thing as the above statement.

This gives you the username that logged in to the session.

These work regardless of sudo or sudo su [whatever]. It also works regardless of how many times su and sudo are called.

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I think $SUDO_USER is valid.

#!/bin/bash
echo $SUDO_USER
whoami

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Here is how to get the username of the person who called the script no matter if sudo or not:

if [ $SUDO_USER ]; then user=$SUDO_USER; else user=`whoami`; fi

or a shorter version

[ $SUDO_USER ] && user=$SUDO_USER || user=`whoami`

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Using whoami, who am i, who, id or $SUDO_USER isn't right here.

Actually, who is never a solution to the question, as it will only list the logged in users, which might be dozens...

In my eyes, the only valuable answer is the use of logname.

Hope this helps

Rob

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If it's the UID you're looking for (useful for docker shenanigans), then this works:

LOCAL_USER_ID=$(id -u $(logname))

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who am i | awk '{print $1}' didn't work for me but who|awk '{print $1}' will serve the job

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Odd, the system does distinguish between real and effective UIDs, but I can find no program that exports this at shell level.