Anchor(faq18)

How can I use numbers with leading zeros in a loop, e.g. 01, 02?

As always, there are different ways to solve the problem, each with its own advantages and disadvantages.

If there are not many numbers, BraceExpansion can be used: 

    for i in 0{1,2,3,4,5,6,7,8,9} 10
    do
        echo $i
    done

Output: 

00
01
02
03
[...]

This gets tedious for large sequences, but there are other ways, too. If the command seq is available, you can use it as follows: 

    seq -w 1 10

or, for arbitrary numbers of leading zeros (here: 3): 

    seq -f "%03g" 1 10

If you have the printf command (which is a Bash builtin, and is also POSIX standard), it can be used to format a number, too: 

    for ((i=1; i<=10; i++))
    do
        printf "%02d " "$i"
    done

The KornShell and KornShell93 have the typeset command to specify the number of leading zeros: 

    $ typeset -Z3 i=4
    $ echo $i
    004

Finally, the following example works with any BourneShell derived shell to zero-pad each line to three bytes: 

i=0
while test $i -le 10
do
    echo "00$i"
    i=`expr $i + 1`
done |
    sed 's/.*\(...\)$/\1/g'

In this example, the number of '.' inside the parentheses in the sed statement determins how many total bytes from the echo command (at the end of each line) will be kept and printed.

One more addendum: in Bash 3, you can use: 

printf "%03d \n" {1..300}

Which is slightly easier in some cases.

Also you can use the printf command with xargs and wget to fetch files: 

printf "%03d \n" {$START..$END} | xargs -i% wget $LOCATION/%

Sometimes a good solution.

I found that on bash 2 you can nest seq in back ticks and this will work as well. 

printf "%03d \n" `seq 300`