Shell Programming and Scripting

hello everyone.. as i said in a previous thread again, im new in linux shell programming and i came up with a new problem i cant solve. so i need your help..

this is the code

Code:

#!/bin/sh
a=0
while [ "$a" -lt 100 ]
do
a=`expr $a + 1`
b=$(( 3 * $a ))
if [ "$a" == "$b" ] ; then
continue
fi
echo "$a"
done

what im trying to do here is display the numbers 1 2 4 5 7 8 10 11 14 15 ....
my code isnt working correctly.. when i run it like this,i get all the numbers displayed even those i dont want (3 6 9 12 etc) and i also get the message unexpected operator after every number..

and when i run it like this

Code:

#!/bin/sh
a=0
while [ "$a" -lt 100 ]
do
a=`expr $a + 1`
b=$(( 3 * $a ))
if [ $a -eq $b ] ; then
continue
fi
echo "$a"
done

i just have all the numbers displayed.. 1 2 3 4 5 6 7 8 9 10... even those i dont want.. so the code is not working correctly..
where is the mistake? im pulling my hair off!

If you are looking to drop multiples of three (you list 15 in your example of desired output, so I'm not 100% sure that's what you want), then what you are comparing is in error.


You are comparing your index with three times the index, and they will never be equal, so you will never skip your printing.


Try something like this:

Code:

#!/bin/sh
      a=0
     while [  "$a" -lt 100 ]

     do
      a=`expr $a + 1`
      b=$((  $a % 3 ))       # if a is evenly divisible by 3, b will be 0

       if [ $b -eq 0 ] ; then
        continue
      fi
      echo "$a"
      done

thank you very much.. this is actually what i had to do from the beggining..
i just want to add that this can also work with

Code:

b=`expr $a % 3`