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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)
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
echo {08..10} solution, however, it is only introduced for bash version 4 and up.
perl) a leading zero indicates an octal number. I suppose it is not often you want a sequence of octal numbers...
seq with the -w switch like so: seq -w 003 produces the numbers 001, 002, and 003.
foo_00001.someextension where nearly all sorting contexts provide a sane ordering.
Jan 16, 2017 at 15:11
I have the same bash version of the original poster (GNU bash, version 3.2.51(1)-release) and {08..12} doesn't work for me, but the following does:
for i in 0{8..9} {10..12}; do echo $i; done
08
09
10
11
12
It's a little more tedious, but it does work on the OP version of bash (and later versions, I would assume).
if you use printf
printf "%.2d " {8..10}
this will force to be 2 chars and will add a leading 0. In case you need 3 digits you can change to "%.3d ".
printf solution, but it is not really making this easier. I am hoping that there is some hidden trick that does the job :)
Use a range that starts with a constant digit and strip that digit off:
echo \ {108..110} | sed 's/ 1//g'
or without using an external command:
a=({108..110}); echo "${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
for loop anyhow.
For reference, I think automatic fixed-width only occurs with the latest version of bash:
$ bash -version
GNU bash, version 3.2.25(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2005 Free Software Foundation, Inc.
$ echo test{00..05}
test0 test1 test2 test3 test4 test5
for i in $(seq -s " " -f %0${zeros}g $ini $fin); do ....;done
zeros, ini and fin are variables, so its easy to fit to your needs. It pads the digits to $zeros zeros from $ini to $fin without problems ;)
