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Korn Shell Loop Problems

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# 1  
Old 01-05-2014
BSD Korn Shell Loop Problems
Very new to the Korn Shell, but I've been looking up loops online and it seems this should work. I'm just trying to convert an ip range in variables $A and $B and iterate the individual ip's out to new lines. Unfortunately I get {152..155} instead of 152, 153, 154, and 155.

Code:
# for i in {$A..$B}; do echo $C.$i; done 
219.19.188.{152..155}

Any tips for a BSD noobie?
I'm using OpenBSD if that helps.
# 2  
Old 01-05-2014
You cannot perform a variable expansion inside a sequence expression.

Another approach would be using seq
Code:
for i in $( seq ${A} ${B} ); do echo $C.$i; done

This User Gave Thanks to Yoda For This Post:
rbatte1
# 3  
Old 01-06-2014
I actually tried that too but it appears seq is not installed on my system:

Code:
# for i in $(seq $A $B); do echo $C.$i; done 
ksh: seq: not found

I tried adding it with pkg_add:

Code:
# pkg_add -r seq
Can't find seq

Do you know if the package for this is called something different?

---------- Post updated 01-06-14 at 12:11 AM ---------- Previous update was 01-05-14 at 11:37 PM ----------

I thought coreutils was the package that contained seq, but I'm still finding this is not installed. pkg_add can't find sysutils either...
# 4  
Old 01-06-2014
Use a while loop instead:
Code:
while [ $A -le $B ]
do
        print $C.$A
        (( A++ ))
done

This User Gave Thanks to Yoda For This Post:
Azrael
# 5  
Old 01-06-2014
I don't doubt this works on your system, but I'm getting some strange output for that?

Code:
# echo $A 
152
# echo $B
155
# while [ $A -le $B ]; do echo $C.$i; ((A++)); done 
219.19.188.10
219.19.188.10
219.19.188.10
219.19.188.10

---------- Post updated at 12:36 AM ---------- Previous update was at 12:27 AM ----------

Got it! I didn't see to change $C.$i to $C.$A. Thanks again Yoda!!!
# 6  
Old 01-06-2014
ksh in OpenBSD is pdksh, which is mostly ksh88, which does not support the brace expansion syntax you were attempting to use.

BSD systems typically have jot instead of seq.

Regards,
Alister
This User Gave Thanks to alister For This Post:
rbatte1
# 7  
Old 01-06-2014
To add to what alister mentioned, you could probably install a valid ksh93 version on your system(s).

--
Quote:
Originally Posted by Yoda
You cannot perform a variable expansion inside a sequence expression.
[..]
Yes you can in ksh, the parsing order for brace expansion is different than in bash.

Quote:
Originally Posted by Yoda
Use a while loop instead:
Code:
while [ $A -le $B ]
do
        print $C.$A
        (( A++ ))
done

If you use arithmetic evaluation with (( .. )), then why not use a for loop:
Code:
for (( i=$A; i<=$B; i++ )); do echo "$C.$i"; done

Or to keep it POSIXy:
Code:
i=$A
while [ $i -le $B ]; do
  printf "%s\n" "$C.$i"
  i=$((i+1))
done

Last edited by Scrutinizer; 01-06-2014 at 04:05 AM..
This User Gave Thanks to Scrutinizer For This Post:
rbatte1
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