You can use the shift example below to get rid of the first parameter. Then at that point the $# will give you the number of parameters left. Then you can loop thru the remaining parameters $1, $2, ... to the second to last one $# - 1. That should give you what you want.
I am no script expert. I just got the idea from a book I bought (recommended on this list). "Portable Shell Programming" by Bruce Blin. It is a good and easy read with plenty of explained examples. Good luck Terry -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Todd A. Jacobs Sent: Thursday, March 13, 2003 3:41 PM To: [EMAIL PROTECTED] Subject: Re: a question about shell script On Thu, 13 Mar 2003, Jihuang Zhou wrote: > Anybody knows how to get an input parameter list from the second > parameter to the second last? For example, if $*="aa bb cc ....mm nn", Why not just shift away the first parameter? foo=$1 shift bar="$*" -- Guvf gntyvar jnf rapbqrq jvgu gur ebg13.fu fpevcg, ninvynoyr ng uggc://jjj.pbqrtabzr.bet/fpevcgvat/fubjfpevcg.cuc?fpevcg=ebg13.fu be sebz n furyy-fpevcgvat nepuvir arne lbh! - ROT-13 Encoded Message -- redhat-list mailing list unsubscribe mailto:[EMAIL PROTECTED] https://listman.redhat.com/mailman/listinfo/redhat-list
