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Top Forums Shell Programming and Scripting Help with IF statement with loop Shell Script Post 302456727 by frans on Saturday 25th of September 2010 01:43:33 PM
Old 09-25-2010
ksh code:
  1. NUMBER=$1
  2. # One argument must be provided, otherwise don't execute
  3. if [ "$#" != 1 ] # $# gives the nb of arguments. There MUST be space after [ and before ]
  4. then
  5.    echo “error: program must be executed with 1 argument.”
  6.    exit
  7. fi # if You already exited, ther's no need to put any else statement
  8. while [ $NUMBER -gt 0 ]
  9. do
  10.    echo -n $NUMBER
  11.    if [ $NUMBER -gt 1 ] # Take care of spaces
  12.    then
  13.       echo -n ", "
  14.    fi
  15.    NUMBER=$(($NUMBER - 1)) # ((NUMBER++)) should work
  16. done
  17. echo
I replaced the printf by echo which is builtin and there's no need of specific formatting.
Just another way to do the same (FYI)
ksh code:
  1. [ "$#" != 1 ] && { echo “error: program must be executed with 1 argument.”; exit; }
  2. for ((i=$1; i>0; i--))
  3. do
  4.    echo -n $i
  5.    ((i>1)) && echo -n ", "
  6. done
  7. echo
 

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