getting user id (owner) of a file / directory 1

[philipose]

hi gurus,

I can get the user id of a file / directory in perl using the below snippet.

$uid = (stat("$pgm_name"))[4];
chomp($userid = qx!cat /etc/passwd | cut -f 1-4 -d ":" | grep -i ":$uid:" | cut -f 1 -d ":"!);
printf "%s\n", $userid;


I need to get the user id in a kshell script. Something like the below
USERID=`perl -e "XXXXXX"`

Any suggestions / alternatives ?
Thanks
philipose

 

[philipose]

I got the below to work but still think that the perl soln was more elegant

ls -ld $DIR} | cut -c15-24 | sed -e 's/ //g'

 

[olded]

It's more efficient to use cut or awk on the 3rd field:

DIR=./mydir

USERID=$(ls -ld "$DIR"|cut -d' ' -f3)
echo $USERID

Some versions of Unix (like Solaris 9) has problems with cut-s -f option. I would use awk:

USERID=$(ls -ld "$DIR"|awk ' { print $3 } ')
echo $USERID

 

[philipose]

Hi olded,
I had also tried to cut the third field but did not work. hence i resorted to cutting the character. the solution you provided with awk works as advertised.
thanks
philipose