Incrementing 001 value in shell script

RobertoBech · 08-20-2007, 10:07 AM

Hi everybody. I'm a beginner in shell-scripting, and I'm stuck at a certain point...

I know how to increment by one the value of a variable. here's an example:

count=1
((count++))
echo $count
2

That's ok. But suppose I have a variable like this:

count=001
((count++))
echo $count
2

That's not what I want at all... I want $count to be 002, not 2. How do I solve this? Thank you all.

unSpawn · 08-20-2007, 10:24 AM

That's *presentation*:

Code:

case ${#count} in 1) count="00${count}";; 2) count="0${count}";; *) ;; esac; echo "$count"

Hko · 08-20-2007, 10:33 AM

or, simpler:

Code:

printf '%0.3d\n' $count

and if you want it to have more digits, just increase the '3' in the above command.

pixellany · 08-20-2007, 10:41 AM

fundamentally, the numbers 002 and 2 are the same. Why 002 gets stored differently I don't know. (Maybe because numbers are stored as strings...)

for example:

Code:

sh-3.1$ a=002
sh-3.1$ echo $a
002
sh-3.1$ b=3
sh-3.1$ echo $b
3
sh-3.1$ echo $((a+b))
5

colucix · 08-20-2007, 11:56 AM

or, another way:

Code:

> count=1001
> ((count++))
> echo ${count:1}
002

RobertoBech · 08-20-2007, 03:55 PM

Quote:

Originally Posted by colucix

or, another way:

Code:

> count=1001
> ((count++))
> echo ${count:1}
002

Thank you all for your suggestions. I've found this one to be the most interesting. Thanks Colucix.

chrism01 · 08-20-2007, 07:38 PM

Pixellany, the 'values' are stored as assigned, but as soon as you do a numeric operation, they get treated as nums (cast to int), which as you know, ignores leading zeroes.