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In the Unix bash shell script, how to capture who the user running the script is?

Hi, in the Unix bash shell script I am trying to capture who the user running the script is.

Question:
What would be a Unix bash shell script/commands so that "me007" (Unix user name) can be replaced with a name of the user running this script?

By the way, this script is called from the .bash_profile

Below is a working Unix script which is intended to delete those unneeded SAS* directories without asking a user too many questions but all files in those directories will be deleted automatically as well. I would like any user to be able to run this script from their .bash_profile file but as you see right now it is not possible to do that because there is a specific user name (me007) hardcoded in the script.

Note:
me007 - a user name on Unix server.
/directory1/tmp - a directory where all these processes will occur.

for dir in $(find /directory1/tmp -type d -a -user me007 -a  -name "SAS*")
do
   echo "Remove $dir (y/n)?  "; read YN
   [[ $YN = Y || $YN = y ]] && rm -r $dir 
done

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SOLUTION

AnthonyHamon

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ASKER CERTIFIED SOLUTION

n2fc

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labradorchik

ASKER

Sorry, but I am not sure if I understand you correctly. How can $LOGNAME be implemented within this script?

Should I just place export $LOGNAME before this script and that is all?
In that case, what should I exchange "me007" with?

n2fc

Try this script to see how the variables react in YOUR system!

echo " LOGNAME: "
echo $LOGNAME
echo "*********"
echo " USER: "
echo $USER
echo "*********"

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n2fc

The export may or may not be necessary...
If needed, it will be the first line in the script...

Try the script sample above... Your system will react properly either with $LOGNAME or $USER instead of "me007"...

labradorchik

ASKER

Oh, just like this?

export $LOGNAME

for dir in $(find /directory1/tmp -type d -a user $LOGNAME -a  -name "SAS*")
do
   echo "Remove $dir (y/n)?  "; read YN
   [[ $YN = Y || $YN = y ]] && rm -r $dir 
done

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n2fc

You forgot - before -user $LOGNAME...

Otherwise OK...
Try each way... $USER and $LOGNAME

SOLUTION

woolmilkporc

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Gerwin Jansen

change the username to:
`whoami`

labradorchik

ASKER

Thank you everyone for your comments and suggestions!!
I just replaced user with $LOGNAME and it worked fine! :)