[SOLVED] print counting using for loop

aggrishabh · 01-11-2012, 02:18 AM

Hi All,

i need to print countiong using for loop like below

001
002
003
004
.
.
.
009
010
.
.
.
999

i tried

Code:

for i in {001..999}
do
print $i
done

but its printing in simple format
1
2
3
.
.
.
9
10
.
.
.

may me the simple one. please help

fukawi1 · 01-11-2012, 02:28 AM

printf will allow you to do it.

Code:

    for ((i=1; i<=999; i++))
    do
        printf "%03d \n" "$i"
    done

"%d" will display a decimal integer, specified by the variable $i. The "0", is the character to lead with, and the 3 is the number of leading characters.

suicidaleggroll · 01-11-2012, 02:46 AM

not sure what language this is, but in bash you could do

Code:

for i in `seq -f "%03g" 1 999`
do
   echo $i
done

David the H. · 01-11-2012, 10:20 AM

Zero-padding in brace expansion was introduced in bash version 4. But since you used "print" and not "echo" in your example, it appears that you may be using ksh or another shell instead (and that's something you really should mention when you post). I'm not sure about the status in those other shells.

For more on zero-padding in bash, see here. It also includes a brief mention about using ksh's typeset.

http://mywiki.wooledge.org/BashFAQ/018

aggrishabh · 01-15-2012, 09:33 AM

Thank you all for your valuable inputs.