Creating string of repeated characters in shell script

Question

I need to generate a string of dots (.characters) as a variable.

I.e., in my Bash script, for input 15 I need to generate this string of length 15: ...............

I need to do so variably. I tried using this as a base (from Unix.com):

for i in {1..100};do printf "%s" "#";done;printf "\n"

But how do I get the 100 to be a variable?

Duplicates: http://stackoverflow.com/q/5349718/2157640 http://stackoverflow.com/q/5799303/2157640 — Palec, Nov 16 '14 at 14:40
n=15; chr='.'; txt=''; for((i=0; $i<$n; i++)); do txt=$txt$chr; done; echo "$txt" — user1855805, Jan 16 '22 at 15:30
This is so badly needed when I use zsh `bindkey` with `^[[D` to move the cursor back into a useful place! — Sridhar Sarnobat, Jun 07 '22 at 23:05

score 85 · Answer 1 · answered May 06 '11 at 19:05

85

You can get as many NULL bytes as you want from /dev/zero. You can then turn these into other characters. The following prints 16 lowercase a's

head -c 16 < /dev/zero | tr '\0' '\141'

answered May 06 '11 at 19:05

Electron

1,390
10
8

10

+1 for a sweet and portable solution without filthy bashisms or hundreds of processes being spawned – fstd May 06 '14 at 02:08
7

you can use /dev/zero directly, no need to redirect – edoceo Jul 20 '15 at 05:41
3

Not *that* sweet and portable: "head: illegal option -- c" (SunOS 5.10 Generic_150401-20 i86pc i386 i86pc) – dokaspar Nov 06 '15 at 08:21
2

`head -c` is portable only on GNU and BSD utils, it's not POSIX. – tricasse Jan 26 '16 at 19:40
3

You can use `dd` instead with `dd if=/dev/zero bs=1 count=16` replacind the `head` part. – NeonMan May 31 '19 at 09:50
`head -n` is [POSIX](https://pubs.opengroup.org/onlinepubs/9699919799/utilities/head.html) – Clint Pachl Mar 26 '23 at 20:25

score 42 · Answer 2 · edited Apr 10 '14 at 11:11

42

len=100 ch='#'
printf '%*s' "$len" | tr ' ' "$ch"

edited Apr 10 '14 at 11:11

Ciro Santilli OurBigBook.com

347,512
102
1,199
985

answered Jul 09 '10 at 12:51

Chris Johnsen

214,407
26
209
186

1

This is the shortest [POSIX 7](http://pubs.opengroup.org/onlinepubs/9699919799/utilities/printf.html#tag_20_94) solution I have seen so far. – Ciro Santilli OurBigBook.com Apr 10 '14 at 11:15
1

can be even shorter, take out the quotation in "$len" to be $len ;) – Sergio Abreu Jan 31 '17 at 15:47
1

`shellcheck` says you are risking code injection without the quotes on `"$len"`. – Jesse Chisholm Feb 15 '22 at 21:41
@JesseChisholm using `shellcheck` you are risking corporate code into someone else's hands ;) – Ricky Levi Apr 19 '23 at 10:41

score 27 · Answer 3 · edited May 23 '17 at 10:31

27

Easiest and shortest way without a loop

VAR=15

Prints as many dots as VAR says (change the first dot to any other character if you like):

printf '.%.0s' {1..$VAR}

Saves the dotted line in a variable to be used later:

line=`printf '.%.0s' {1..$VAR}`
echo "Sign here $line"

-Blatantly stolen from dogbane's answer https://stackoverflow.com/a/5349842/3319298

Edit: Since I have now switched to fish shell, here is a function defined in config.fish that does this with convenience in that shell:

function line -a char -a length
  printf '%*s\n' $length "" | tr ' ' $char
end

Usage: line = 8 produces ========, line \" 8 produces """""""".

edited May 23 '17 at 10:31

Community

1
1

answered Apr 15 '14 at 15:04

Epigene

3,634
1
26
31

1

what an interesting way to exploit printf's shell feature that repeatibly consumes arguments until satisfied in combination with a zero-field width, forcing just the string before the % to be printed. This could be used as a trick to divide the count as well (adding the %.0s twice to force it to consume 2 arguments per pass) – osirisgothra Jun 22 '14 at 18:42
oh yeah and I forgot something important, since printf processes %s with a null if no args are given, a value of zero would be a problem, for that you would have to add an if statement – osirisgothra Jun 22 '14 at 21:02
3

This only works for me in my shell (bash) if I use eval: `VAR=15 ; eval printf '=%.0s' {1..$VAR}` – 6EQUJ5 Sep 27 '14 at 06:42
Only works for me with eval too. Replacing $VAR with $COLUMNS is very useful for making a horizontal ruler. If needed from inside a script: `hr () { eval printf '=%.0s' {1..$(tput cols)}; echo; }` – Nov 28 '14 at 09:37
Your answer is fine, using *POSIX shell*, but using [tag:bash] , you could avoid one fork by using `-v` arg to `printf -v line '.%.0s' {1..$VAR}` – F. Hauri - Give Up GitHub Mar 20 '17 at 07:41
6

@6EQUJ5 In `bash`, you can't use variables in a brace expansion. Instead, you can use `$(seq 1 $VAR)` in place of `{1..$VAR}` – wisbucky Dec 08 '17 at 23:56
@wisbucky actually it does... tested exactly what I quoted above just now using bash version 4.3.30(1)-release on Debian 8 - maybe you have a different shell ```~$ VAR=15 ; eval printf '=%.0s' {1..$VAR}``` (line break was here) ```===============``` – 6EQUJ5 Dec 11 '17 at 09:52
1

@6EQUJ5 I meant without using `eval`. `eval` is not safe to use in a script. https://stackoverflow.com/questions/17529220/why-should-eval-be-avoided-in-bash-and-what-should-i-use-instead – wisbucky Dec 11 '17 at 18:32

score 9 · Answer 4 · answered Jul 09 '10 at 11:25

9

On most systems, you could get away with a simple

N=100
myvar=`perl -e "print '.' x $N;"`

answered Jul 09 '10 at 11:25

Christopher Creutzig

8,656
35
45

score 6 · Answer 5 · edited May 23 '17 at 12:26

I demonstrated a way to accomplish this task with a single command in another question, assuming it's a fixed number of characters to be produced.

I added an addendum to the end about producing a variable number of repeated characters, which is what you asked for, so my previous answer is relevant here:

https://stackoverflow.com/a/17030976/2284005

I provided a full explanation of how it works there. Here I'll just add the code to accomplish what you're asking for:

    n=20 # This the number of characters you want to produce

    variable=$(printf "%0.s." $(seq 1 $n)) # Fill $variable with $n periods

    echo $variable # Output content of $variable to terminal

Outputs:

....................

n=0 fails to produce the correct number of characters. As do most of the printf solutions. — NOYB, Sep 07 '20 at 19:19

score 4 · Answer 6 · answered Jul 09 '10 at 15:16

You can use C-style for loops in Bash:

num=100
string=$(for ((i=1; i<=$num; i++));do printf "%s" "#";done;printf "\n")

Or without a loop, using printf without using any externals such as sed or tr:

num=100
printf -v string "%*s" $num ' ' '' $'\n'
string=${string// /#}

Aleksey Otrubennikov · Answer 7 · 2010-07-09T12:38:23.270

3

The solution without loops:

N=100
myvar=`seq 1 $N | sed 's/.*/./' | tr -d '\n'`

edited Jul 09 '10 at 12:38

answered Jul 09 '10 at 12:04

Aleksey Otrubennikov

1,121
1
12
26

score 2 · Answer 8 · answered Feb 10 '14 at 17:13

2

num=100
myvar=$(jot -b . -s '' $num)
echo $myvar

answered Feb 10 '14 at 17:13

Alexander Gladysh

39,865
32
103
160

score -1 · Answer 9 · answered Jul 09 '10 at 14:50

-1

When I have to create a string that contains $x repetitions of a known character with $x below a constant value, I use this idiom:

base='....................'
# 0 <= $x <= ${#base}
x=5
expr "x$base" : "x\(.\{$x\}\)"    # Will output '\n' too

Output:

.....

answered Jul 09 '10 at 14:50

dolmen

8,126
5
40
42

2

In bash and ksh (perhaps zsh), you could write `echo ${base:0:$x}` -- details [here](http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion) – glenn jackman Jul 22 '14 at 13:21

Creating string of repeated characters in shell script

9 Answers9

Linked