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Home of user that is stored in var

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# 1  
Old 10-26-2010
Home of user that is stored in var
I have a user name that is stored in variable $i
and i want to use that user's home dirctor in case command

something like this


find ~"$i" |while read p
do
case "$p" in
( ~"$i"/myDir )
echo "$p"
;;
(*)
esac
done

but it doesn't work

some help please
Last edited by radoulov; 10-26-2010 at 12:36 PM.. Reason: Code tags, please!
# 2  
Old 10-26-2010
Please define "doesn't work" like error messages or unexpected behavior.
Your script sample works fine with ksh.
# 3  
Old 10-26-2010
In ksh, eval runs your line through the ksh one more time:
Code:
$( eval echo ~$i )

or

$( echo echo ~$i | ksh )

or in your case:

eval find ~$i . . . | . . . .

You can go get it yourself from /etc/passwd's final field.
Last edited by DGPickett; 10-26-2010 at 12:48 PM..
This User Gave Thanks to DGPickett For This Post:
testman84
# 4  
Old 10-26-2010
it show :
find: '~jan44' : no such file or director

where jan44 is the value of the $i
and i have a user with name jan44 on my system " debian" and i use bash
# 5  
Old 10-26-2010
Eval isn't required with ksh but will fix this script when run by bash:
Code:
ihome=$(eval echo ~$i)
find $ihome |while read p
do
  case "$p" in
  ($ihome/myDir)
    echo "$p"
  ;;
  esac
done

This User Gave Thanks to jlliagre For This Post:
testman84
# 6  
Old 10-26-2010
it works Thanks
# 7  
Old 10-26-2010
Maybe I have an old ksh? Revision: 82.10.1.61

You can do the eval right on the find, saving much clutter.
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