33

In zsh, when I have to create a bunch of files with zsh, I usually do something like:

for x in $(seq 1 1000); do .....; done

This works fine, it gives me files with names foo-1.txt .. foo-1000.txt.
However, these files do not sort nicely, so I would like to zero-pad the $x variable, thus producing names of foo-0001.txt .. foo-1000.txt.

How to do that in zsh? (and bonus question, how to do it in bash?)

Mat
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ervingsb
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6 Answers6

35

For reference sake, if you do not have control over the generation of the numeric values, you can do padding with:

% value=314
% echo ${(l:10::0:)value}
0000000314
% echo $value
314
Francisco
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    Seeing that ${(l... is not very search-engine friendly, look this up under Parameter Expansion Flags in zshexpn(1). Also paired delimiters can be used to make the syntax a bit more intuitive: ${(l{10}{0})value}. – Roman Odaisky Jul 22 '13 at 20:57
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    Strangely, zmv doesn’t understand that form, though parentheses work fine: ${(l(10)(0))value}. – Roman Odaisky Jul 22 '13 at 21:11
  • echo ${(l:3::0:)$((++value))} – zzapper Jan 27 '22 at 13:39
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    a strange caveat: if the value is longer than the given length the first digits are cut of, e.g. for l=3 and a value of 1234 the output is 234. – A. Rabus Jul 29 '22 at 07:50
19

Use the -w flag to seq (in any shell):

$ seq -w 1 10
01
02
03
04
05
06
07
08
09
10
Mat
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12

You can use bash's brace expansion:

$ for n in file-{0001..1000}.txt; do echo $n; done
file-0001.txt
file-0002.txt
file-0003.txt
file-0004.txt
file-0005.txt
file-0006.txt
file-0007.txt
file-0008.txt
file-0009.txt
file-0010.txt
...
file-0998.txt
file-0999.txt
file-1000.txt

Works in zsh too.

glenn jackman
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9

in Zsh:

% my_num=43
% echo ${(l(5)(0))my_num}
00043
Konstantin Nikitin
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2

Works in bash (don't know for zsh):

echo foo{0000..1000}.txt
user unknown
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1

  1. printf
    Using printf to pad with zeros works in any shell:

    $ printf 'foo%03d.txt\n' $(seq 4)
foo001.txt
foo002.txt
foo003.txt
foo004.txt

$           printf  'foo%03d.txt\n' {1..4}       # works in bash, ksh, zsh.
$ n=4;      printf  'foo%03d.txt\n' {1..$n}      # only ksh, zsh
$ n=4; eval printf '"foo%03d.txt\n" {1..'"$n"'}' # workaround needed for bash

  2. Use brace expansion (only bash and zsh keep the leading zeros ksh don't).

    $ printf '%s\n' foo{1..004}.txt
foo001.txt
foo002.txt
foo003.txt
foo004.txt