Shell Scripting

Need help please - new to UNIX. Am trying to write a script that does several things but am having trouble with this section: I have a directory and files listed on the command line (i.e. runthis.sh /export/home/myhome file1 file2 )

If file1 is found, I want to display the full path on the screen.

I have beent trying to use the while...do by setting $kount and checking if $kount is gt $#. That seems to work but I can not get the $2, $3, $4.... set using $"$kount". When I echo $"$kount" it displays $2 but when I use it to find the file, I get an error message.

Thanks for any and all help.

Bobby

We can not get argument parameters the manner you have tried there. We have to transfer into an array and get those variables.

$# -- will give number of argument variables
$* -- will give all argument variables
$@ -- will give all argument variables and equivalent to "$*"
see man ksh for this.

We can try with an example as,

#!/usr/bin/sh
# Debugging mode
set -x

# Array transmission
VAR=($*)

i=0

while [[ $i -lt ${#VAR[*]} ]]
do

echo ${VAR[$i]}
let i=i+1
done

Sample execution gives as,

$ sh test.sh /export/home/myhome file1 file2
+ VAR=($*)
+ i=0
+ [[ 0 -lt 3 ]]
+ echo /export/home/myhome
/export/home/myhome
+ let i=i+1
+ [[ 1 -lt 3 ]]
+ echo file1
file1
+ let i=i+1
+ [[ 2 -lt 3 ]]
+ echo file2
file2
+ let i=i+1
+ [[ 3 -lt 3 ]]


VAR[0] -- will be with /export/home/myhome
VAR[1] -- with file1
VAR[2] -- with file2

HTH.

REgards,
Muthukumar.

try something like: