How to create a sequence with leading zeroes using brace expansion

Question

When I use the following, I get a result as expected:

$ echo {8..10}
8 9 10

How can I use this brace expansion in an easy way, to get the following output?

$ echo {8..10}
08 09 10

I now that this may be obtained using seq (didn't try), but that is not what I am looking for.

Useful info may be that I am restricted to this bash version. (If you have a zsh solution, but no bash solution, please share as well)

$ bash -version
GNU bash, version 3.2.51(1)-release (x86_64-suse-linux-gnu)

Answer 1

Use a range that starts with a constant digit and strip that digit off:

echo \ {108..110} | sed 's/ 1//g'

or without using an external command:

a=({108..110}); echo "${a[@]#1}"

For use in a for loop:

for x in {108..110}; do
  x=${x#1}
  …
done

though for this case a while loop would be clearer and works in any POSIX shell:

x=8
while [ $x -le 10 ]; do
  n=$((100+x)); n=${n#1}
  …
  x=$((x+1))
done

Answer 2

Prefix the first number with a 0 to force each term to have the same width.

$ echo {08..10}
08 09 10

From the bash man page section on Brace Expansion:

Supplied integers may be prefixed with 0 to force each term to have the same width. When either x or y begins with a zero, the shell attempts to force all generated terms to contain the same number of digits, zero-padding where necessary.

Also note that you can use seq with the -w option to equalize width by padding with leading zeroes:

$ seq -w 8 10
08
09
10

$ seq -s " " -w 8 10
08 09 10

If you want more control, you can even specify a printf style format:

$ seq -s " " -f %02g 8 10
08 09 10

Answer 3

if you use printf

printf "%.2d " {8..10} 

this will force to be 2 chars and will add a leading 0. In case you need 3 digits you can change to "%.3d ".