Hi,

You could use this too

#!/bin/bash # # e.g. sample.sh foo1 foo2 foo3 # then the following will happen : arg[0]=foo1, arg[1]=foo2 , arg[2]=foo3 #

count=0 # Counts the number of parameters

until [ -z "$1" ] # loop until all the parameters are usedup do arg[$count]=$1 # the $arg[] is like a normal array and # $1 is the first argument to the script

echo -en "Argument $count is $arg[$count]\n" # Debug Output

count=$(($count+1))

shift # shifts the argument number by one thus # now $1 is argument number two rather being argument one done

thanks Ripunjay Bararia

-----Original Message----- From: linuxers-admin@mm.ilug-bom.org.in [mailto:linuxers-admin@mm.ilug-bom.org.in]On Behalf Of Nikhil Joshi Sent: Monday, December 16, 2002 9:59 PM To: ilug Subject: [ILUG-BOM] Bash script doubt

**************************************** * LUG meet on 12 Jan. 2003 @ VJTI ****************************************

Hi!

I'm trying to take command line args with the help of a bash script But i'm not able to achieve it Pls guide.

#!/bin/bash # i want to detect command line options # e.g. sample.sh -1 foo1 -2 foo2 -3 foo3 # then i want following : arg1=foo1, arg2=foo2 , arg3=foo3 # I tried following but arg1 takes the value $2, arg2 $3 and arg3 $3

count=0 # counter to track current command line arg

for cmd in $@ do count=`expr $count + 1` # count = count + 1 if `test $cmd = -1` then arg1=$`expr $count + 1` echo $arg1 # unfortunately it gives $2 and not foo1 fi done

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On Mon, 16 Dec 2002, Nikhil Joshi wrote:

# i want to detect command line options # e.g. sample.sh -1 foo1 -2 foo2 -3 foo3 # then i want following : arg1=foo1, arg2=foo2 , arg3=foo3 # I tried following but arg1 takes the value $2, arg2 $3 and arg3 $3

try this one, i tried with getopts, didnt impress me.

#!/bin/bash

i=0 # counter

while test $# -ne 0; do p=$1 # getting parameter value let i=i+1 if test $p = "-${i}"; then shift eval arg$i=$1 # argno=value fi shift done

echo "arg1=$arg1, arg2=$arg2"

regards sanjeev