Creating numerous directories using mkdir

Question

I would like to create many directories using mkdir. Each directory name will consist of a prefix (a string) and an index (an integer). Suppose that I would like the prefix to be "s" and the indices to range from 1 to 50. This means that I would like to create directories titled:

s1, s2, ... , s49, s50

Is there a way to do this automatically using mkdir? Thanks for your time.

Answer 1

• One

  for i in {1..50}; do
    mkdir s"$i"
  done
  

• Two

  mkdir s{1..50}
  

  This option works in bash, zsh and ksh93

• Three

  mkdir $(printf "s%02i " $(seq 1 50))
  

Answer 2

You can do this with a shell script.

Pure sh - this will work even on pre-POSIX bourne shells:

n=1;
max=50;
while [ "$n" -le "$max" ]; do
  mkdir "s$n"
  n=`expr "$n" + 1`;
done

If you want to create a high number of directories, you can make the script faster by reducing it to a single call of mkdir as well as using shell builtins for testing and arithmetics. Like this:

n=1
max=50
set -- # this sets $@ [the argv array] to an empty list.

while [ "$n" -le "$max" ]; do
    set -- "$@" "s$n" # this adds s$n to the end of $@
    n=$(( $n + 1 ));
done 

mkdir "$@"

Zsh, ksh93 or bash make this much easier, but I should point out this is not built into mkdir and may not work in other shells. For larger cases, it may also be affected by limits on the number or total size of arguments that may be passed to a command.

mkdir s{1..50}

Answer 3

Lots of complicated answers here, but bash makes it really easy. Sure, the pure POSIX solution works, but why not take advantage of the bash shell you're using, anyhow? You can do this easily with brace expansion:

% mkdir -v s{1..10} && ls -1d s{1..10}                                   (09-24 17:37)
mkdir: created directory `s1'
mkdir: created directory `s2'
mkdir: created directory `s3'
mkdir: created directory `s4'
mkdir: created directory `s5'
mkdir: created directory `s6'
mkdir: created directory `s7'
mkdir: created directory `s8'
mkdir: created directory `s9'
mkdir: created directory `s10'
s1
s10
s2
s3
s4
s5
s6
s7
s8
s9

Answer 4

mkdir $(seq --format 's%.0f' 1 50)

or if you want zero-padded numbers (which would be better for sorting):

mkdir $(seq --format 's%02.0f' 1 50)

or:

mkdir s$(seq -s ' s' -w 1 50) -- note the string 's' just before the $(), without it the first directory created will be just '01' rather than 's01'

and, finally: mkdir $(printf "s%02i " $(seq 1 50))

seq is from GNU Coreutils

oddly, seq's --format or -f option only allows printf's floating point double types (like f and g. also a strange floating point hex format that i've never found any use for). I have no idea why. It would be nice if it also supported other printf(3) numeric types like integer (d,i), octal (o,U) or hex (x,X).

Anyway, a double format with 0 decimal precision like %.0f or %02.0f is close enough to an integer for this purpose.

$ seq --help
Usage: seq [OPTION]... LAST
  or:  seq [OPTION]... FIRST LAST
  or:  seq [OPTION]... FIRST INCREMENT LAST
Print numbers from FIRST to LAST, in steps of INCREMENT.

  -f, --format=FORMAT      use printf style floating-point FORMAT
  -s, --separator=STRING   use STRING to separate numbers (default: \n)
  -w, --equal-width        equalize width by padding with leading zeroes
      --help     display this help and exit
      --version  output version information and exit

If FIRST or INCREMENT is omitted, it defaults to 1.  That is, an
omitted INCREMENT defaults to 1 even when LAST is smaller than FIRST.
FIRST, INCREMENT, and LAST are interpreted as floating point values.
INCREMENT is usually positive if FIRST is smaller than LAST, and
INCREMENT is usually negative if FIRST is greater than LAST.
FORMAT must be suitable for printing one argument of type `double';
it defaults to %.PRECf if FIRST, INCREMENT, and LAST are all fixed point
decimal numbers with maximum precision PREC, and to %g otherwise.

See also: http://www.gnu.org/software/coreutils/manual/html_node/seq-invocation.html

Answer 5

Just to be different, here's a POSIX sh solution that uses recursion:

makedirs() {
  [ "$1" -gt 0 ] || return
  mkdir "s$1"
  makedirs $(( $1 - 1 ))
}
$ makedirs 9
$ ls
s1  s2  s3  s4  s5  s6  s7  s8  s9