Iterating through a range of ints in ksh?

Question

How can I iterate through a simple range of ints using a for loop in ksh?

For example, my script currently does this...

for i in 1 2 3 4 5 6 7
do
            


        

Answer 1

Just a few examples I use in AIX because there is no range operator or seq, abusing perl instead.

Here's a for loop, using perl like seq:

for X in `perl -e 'print join(" ", 1..10)'` ; do something $X ; done

This is similar, but I prefer while read loops over for. No backticks or issues with spaces.

perl -le 'print "$_ " for 1..10;' | while read X ; do xargs -tn1 ls $X ; done

My fav, do bash-like shell globbing, in this case permutations with perl.

perl -le 'print for glob "e{n,nt,t}{0,1,2,3,4,5}"' | xargs -n1 rmdev -dl

Answer 2

ksh93, Bash and zsh all understand C-like for loop syntax:

for ((i=1; i<=9; i++))
do
    echo $i
done

Unfortunately, while ksh and zsh understand the curly brace range syntax with constants and variables, Bash only handles constants (including Bash 4).

Answer 3

While loop?

while [[ $i -lt 1000 ]] ; do
    # stuff
   (( i += 1 ))
done

Answer 4

Curly brackets?

for i in {1..7}
do
   #stuff
done

Answer 5

seq - but only available on linux.

for i in `seq 1 10`
do 
    echo $i
done

there are other options for seq. But the other solutions are very nice and more important, portable. Thx

Answer 6

on OpenBSD, use jot:

for i in `jot 10`; do echo $i ; done;