UNIX for Dummies Questions & Answers

hi all,

i have a ksh script that takes up to 3 parameters -- only 2 of which are required. what's the simplest way to check if the user passed 2 or 3 parameters?

if 3 parameters are not null then
do this
elif 2 parameters are not null then
do this
else
echo "you need at least 2 parameters"

thanks for you help!

hi again.

maybe i should add that i understand this much:

if [ -n "$1" ]; then

but i'm unclear on how to check for multiple parameters. i've tried different things, searched through this forum and scanned the o'reilly's korn shell book, but cannot find the answer.

thanks.

The variable $# gives the number of the arguments.

Regards

thanks for your response! do you mean something like this? it's not quite working for me. i'm pretty new at this.


if [ $# == 3 ]; then
QUERY = "$QUERY update user set password=password('$2') where user='$1' and host='$3'"
elif [ $# == 2 ]; then
QUERY = "$QUERY update user set password=password('$2') where user='$1'"
else
echo "At a minimum, you must provide a user and password: chpass <user> <new password>"
fi


thanks.

The comparison operator is wrong, it should be -eq for an integer.
Don't use single quotes for the variables, they prevent the shell to expand them.

Code:

if [ $# -eq 3 ]; then
  QUERY = "$QUERY update user set password=password($2) where user=$1 and host=$3"
elif [ $# -eq 2 ]; then
  QUERY = "$QUERY update user set password=password($2) where user=$1"
else
  echo "At a minimum, you must provide a user and password: chpass <user> <new password>"
fi

For basic usage like this you really should have a read of a book or tutorial regarding shell scripting, a link to some tutorials:

https://www.unix.com/answers-frequent...tutorials.html

Regards

Single quotes inside double quotes are not special, so those should be fine.

Let me remind you of the case statement.

Code:

case $# in 2|3) ;; *) echo "Game over, try again?" >&2;; esac