Script: how translate for i in `seq $1 $2`?

[aleunix]

How translate the bash command:
for i in `seq $1 $2`

on:
$ echo $SHELL
/bin/ksh

?

Essentially i would like pass two argument on for instead to put two fix number all the time.

Now i have problem with seq (bash).

[denta]

man jot

[aleunix]

Thanks.

#!/bin/sh
for i in `jot - $2 $3 1`
do
echo $i
done

[IdOp]

Glad the problem is solved with jot. Just to clarify ...

Is seq really a bash command now? I don't find it mentioned in the bash man pages I looked at (though they're not the very latest). Both NetBSD and some Linux distributions have a stand-alone seq(1) command. And bash has a construct (in newer versions anyway)

% echo {1..10}
1 2 3 4 5 6 7 8 9 10

which does a similar thing (in simple cases).

[aleunix]

Now i'm on OpenBSD so i don't check it but i think you have right.

Now i was thinking at something sufficient general for avoid to use seq or jot that can be work on OpenBSD and Linux without modification of the code.
These attempt failed:

for i in {$1.. $2}
for ((i=$1; i <= $2 ; i++))

so, in the end, now i'm using that approach:
i=$1
while [ "$i" -le $2 ]
do
echo $i
let "i+=1"
done
I haven't check on Linux but given that i make this change while see it:
Example 11-17. C-style syntax in a while loop from the Advanced Bash-Scripting Guide
i suppose that should work.

[IdOp]

Originally Posted by aleunix

These attempt failed:
for i in {$1.. $2}

{ .. } is a bash-ism not found in ksh.

for ((i=$1; i <= $2 ; i++))

The (( )) should contain an arithmetic expression, rather than a command list.

now i'm using that approach:
i=$1
while [ "$i" -le $2 ]
do
echo $i
let "i+=1"
done
I haven't check on Linux ...

I checked and it works.