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# 1  
Old 10-19-2007
Scripting Newbie

Seems simple but I am having difficulty with this one: I am trying to write a single command line argument (which will be a path) - the program should print out the owner of the path.
I can not get anything I write to run.
Please help.
# 2  
Old 10-19-2007
Quote:
Originally Posted by Kymmers7
Seems simple but I am having difficulty with this one: I am trying to write a single command line argument (which will be a path) - the program should print out the owner of the path.
I can not get anything I write to run.
Please help.


Sorry, could you please be more descriptive.

To find the owner of the path. -- path could either be a file or a directory.

Did you mean finding the owner of the file or a directory ?
# 3  
Old 10-19-2007
Yes, find owner of file or directory to print that information out once the script is run. I know the command to get the information, I am having difficulty writing the command within a script for a user to supply a file/directory to get that information.
# 4  
Old 10-19-2007
Code:
#! /bin/zsh

if [ $# -eq 0 ]
then
  echo "USAGE: script file/directory"
  exit 1
fi

if [ -f $1 ]
then
  echo "Owner of $1 : `ls -l $1 | awk '{ print $3 }'`"
else
  echo "Owner of $1 : `ls -ld $1 | awk '{ print $3 }'`"
fi

exit 0

# 5  
Old 10-19-2007
Does it matter if it is for the C Shell?
# 6  
Old 10-19-2007
if you have GNU find

Code:
find /path -printf "File:%f - Owner:%u\n"


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