Hi all, I am writing a unix script and am desperately trying to have it accept parameters not using the $1, $2,.. $9 way but by allowing arguments on the command line, e.g. "call_script -name John -age 24" instead of "call_script John 24" in bash 2.03... Can't find out how to do it... but it must be possible... please help... thanx alot!!!

Look at getopt in the bash shell or getopts in ksh.

Here is an example from a script I had: # Use getopts to scan for -e/-w/-p/-f [filename]: # (':' after option letter indicates arg expected. # Leading ':' in optstring suppresses default error messages.) error_level='ERROR' # Default prompt_for_comments='FALSE' # Default while getopts ":ewpf:" opt; do case $opt in e ) error_level='ERROR' ;; w ) error_level='WARNING'; message_prefix='Warning: ' ;; p ) prompt_for_comments='TRUE' ;; f ) filename=$OPTARG ;; * ) print -u2 'fcheck: invalid option.'; exit 1 ;; esac done shift $(($OPTIND -1)) print error_level: $error_level print prompt_for_comments: $prompt_for_comments print filename: $filename Note a couple of limitations though: * You can only parse single-letter flags, e.g. "-a 24", not "-age 24" * If an invalid option is passed, there is no straightforward way AFAIK to display the offending option, since "$opt" just gives "?".

Arne: Take a look at the my_getopts function at: http://www.unixreview.com/documents/s=1344/uni1042138723500/ my_getopts will let you do AGE=24. There's a version for bash as well as ksh. Regards, Nails