bash sequence 00 01 ... 10 - bash

Question

in bash, with
$ echo {1..10}
1 2 3 4 5 6 7 8 9 10
I can get a numbers sequence, but in some case I need
01 02 03 ... 10
how I can get this ?
and how I can get ?
001 002 ... 010 011 .. 100

Answer 1

This will work in any shell on a machine that has coreutils installed (thanks commenters for correcting me):
seq -w 1 10
and
seq -w 1 100
Explanation:
the option -w will:
Equalize the widths of all numbers by padding with zeros as necessary.
seq [-w] [-f format] [-s string] [-t string] [first [incr]] last
prints a sequence of numbers, one per line (default), from
first (default 1), to near last as possible, in increments of incr (default
1). When first is larger than last the default incr is -1

Answer 2

use seq command with -f parameter, try:
seq -f "%02g" 0 10
results:
00
01
02
03
04
05
06
07
08
09
10
seq -f "%03g" 0 10
results:
000
001
002
003
004
005
006
007
008
009
010

Answer 3

printf "%02d " {1..10} ; echo
Output:
01 02 03 04 05 06 07 08 09 10
Similarly:
printf "%03d " {1..100} ; echo
In more recent versions of bash, simply:
echo {01..10}
And:
echo {001..100}

Answer 4

for i in {01..99}; do
echo $i
done
will return :
01
02
03
04
05
06
07
08
09
10
...
Replacing 01 with 001 and 99 with 999 or 100 will do what you expect also.

Answer 5

$ printf "%02d " {0..10}; echo
00 01 02 03 04 05 06 07 08 09 10
$ printf "%03d " {0..100}; echo
000 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 053 054 055 056 057 058 059 060 061 062 063 064 065 066 067 068 069 070 071 072 073 074 075 076 077 078 079 080 081 082 083 084 085 086 087 088 089 090 091 092 093 094 095 096 097 098 099 100
Just vary the field width in the format string (2 and 3 in this case) and of course the brace expansion range. The echo is there just for cosmetic purposes, since the format string does not contain a newline itself.
printf is a shell builtin, but you likely also have a version from coreutils installed, which can be used in-place.

Answer 6

awk only:
awk 'BEGIN { for (i=0; i<10; i++) printf("%02d ", i) }'

Answer 7

The following will work in bash
echo {01..10}
**EDIT seeing the answers around me I just wanted to add this, in the case we're talking about commands that will work under any terminal
yes | head -n 100 | awk '{printf( "%03d ", NR )}' ##for 001...100
or
yes | head -n 10 | awk '{printf( "%03d ", NR )}' ##for 01..10

Answer 8

echo 0{0..9}
You can get: 00 01 02 03 04 05 06 07 08 09
echo 0{0..9} 1{0..9}
You can get: 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19
echo 00{0..9} 0{10..99}
You can get 001 .. 099

Answer 9

There are so many ways to do this! My personal favorite is:
yes | grep y | sed 100q | awk '{printf( "%03d ", NR )}'; echo
Clearly, neither the sed nor the grep are necessary (the grep being far more trivial, since if you omit the sed you need to change the awk), but they contribute to the overall satisfaction of the solution! The final echo is not really necessary either, but it's always nice to have a trailing newline.
Another nice option is:
yes | nl -ba | tr ' ' 0 | sed 100q | cut -b 4-6
Or (less absurdly):
yes '' | sed ${top-100}q | nl -ba -w ${width-3} -n rz

Answer 10

as commented by favoretti, seq is your friend.
But there is a caveat:
seq -w uses the second argument to set the format it will use.
Thus, the command seq -w 1 9 will print the sequence 1 2 3 4 5 6 7 8 9
To print the sequence 01 .. 09 you need to do the following:
seq -w 1 09
Or for clarities sake use the same format on both ends, for instance:
seq -w 000 010 for the series 001 002 003 ... 010
And you can also use a step argument that also works in reverse:
seq -w 10 -1 01' for 10,09,08...01 orseq -w 01 2 10` for 01,03,05,07,09