Creating multiple directories using a parameter in a shell script

Question

I'm trying to make a bash script that would make a series of directories and requesting a parameter of how many directories should be created.

$> ./createDir.sh 5

$> ls

ex_01
ex_02
ex_03
ex_04
ex_05

I tried using mkdir ex_{01..$1} but it does not seem correct. How could I make this work (without using any loop)?

it is ! We haven't seen the loops yet and it should be easier than making loops. Is it me having the syntax wrong or is it not possible to use a parameter in this case? — RayedB, Oct 04 '16 at 12:49
That's zsh syntax, replace #! /bin/bash with #! /bin/zsh — Stéphane Chazelas, Oct 04 '16 at 12:57
Similar: difference between for i in {1..$N} and for i in $(seq 1 1 $N) — Stéphane Chazelas, Oct 04 '16 at 13:00

score 1 · Accepted Answer · edited May 23 '17 at 12:40

1

You will need eval for this.

#!/bin/bash

start=1
stop=$1

mkdir $(eval echo ex_{$start..$stop})

But I agree with don_crissti, why not simply use a loop?

Before:

ls -p | grep 'ex_'
<empty>

After I run the script:

./makeDirs.sh 3
ls -p | grep 'ex_'
ex_1/
ex_2/
ex_3/

Further reading:

edited May 23 '17 at 12:40

Community

answered Oct 04 '16 at 12:50

Good on including the "why is eval bad" link. – Wildcard Oct 04 '16 at 12:57
2

No need for echo. eval "mkdir ex_{$start..$stop}" should be enough. – Stéphane Chazelas Oct 04 '16 at 13:04
@StéphaneChazelas Yes, you're right. Works without echo as well. – Oct 04 '16 at 15:35

1 Answers1