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This gets tedious for large sequences, but there are other ways, too.
If you have the {{{printf}}} command (which is a Bash builtin, and is also POSIX standard), it can be used to format a number:
This gets tedious for large sequences, but there are other ways, too. If you have the {{{printf}}} command (which is a Bash builtin, and is also POSIX standard), it can be used to format a number:
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    for ((i=1; i<=10; i++)) # Bash 2 for-loop syntax     for ((i=1; i<=10; i++))     # Bash 2 for-loop syntax
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In Bash 3, you can use ranges inside brace expansion.
Also, since {{{printf}}} will implicitly loop if given more arguments than format specifiers, you can simplify this enormously:
In Bash 3, you can use ranges inside brace expansion. Also, since {{{printf}}} will implicitly loop if given more arguments than format specifiers, you can simplify this enormously:
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   printf "%03d\n" {1..300} # Bash 3 brace expansion    printf "%03d\n" {1..300}     # Bash 3 brace expansion
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(That may be helpful if your version of {{{seq(1)}}} lacks {{{printf}}}-style format specifiers. Since it's a nonstandard external tool, it's good to keep your options open.)
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(That may be helpful if your version of {{{seq(1)}}} lacks {{{printf}}}-style format specifiers. Since it's a nonstandard external tool, it's good to keep your options open.) Be warned however that seq might be considered bad style, it's even mentioned in ["Don't Ever Do These"].
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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 you have the printf command (which is a Bash builtin, and is also POSIX standard), it can be used to format a number:

    for ((i=1; i<=10; i++))     # Bash 2 for-loop syntax
    do
        printf "%02d " "$i"
    done

In Bash 3, you can use ranges inside brace expansion. Also, since printf will implicitly loop if given more arguments than format specifiers, you can simplify this enormously:

   printf "%03d\n" {1..300}     # Bash 3 brace expansion

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

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

If the command seq(1) is available (it's part of GNU sh-utils/coreutils), you can use it as follows:

    seq -w 1 10

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

    seq -f "%03g" 1 10

Combining printf with seq(1), you can do things like this:

   printf "%03d\n" $(seq 300)

(That may be helpful if your version of seq(1) lacks printf-style format specifiers. Since it's a nonstandard external tool, it's good to keep your options open.)

Be warned however that seq might be considered bad style, it's even mentioned in ["Don't Ever Do These"].

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 command determines how many total bytes from the echo command (at the end of each line) will be kept and printed.

Now, since the number one reason this question is asked is for downloading images in bulk, you can use the printf command with xargs(1) and wget(1) to fetch files:

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

Or, in a slightly more general case:

   for i in {1..100}; do
      wget "$prefix$(printf %03d $i).jpg"
      # other commands
   done

BashFAQ/018 (last edited 2019-08-21 16:24:29 by GreyCat)