shell script create string of repeated characters

Question

This seems like a simple problem but I couldnt find a ready solution. I need to generate a string of characters "."s as a variable.

ie, IN my bash script, I need to generate a string of length 15 ...............

I need to do so variably. I tried using this as a base (from: http://www.unix.com/shell-programming-scripting/46584-repeat-character-printf.html)

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

But how do i get the 100 to be a vairbale?

Answer 1

I don't know if I got the question.. maybe you can do it like this

function myPrint { for i in `seq 1 $2`; do echo -n $1; done }; myPrint <char> <number>

Answer 2

Just put your code into $( )

myvar=$(for i in {1..100};do printf "%s" "#";done;printf "\n")

Answer 3

On most systems, you could get away with a simple

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

Answer 4

The solution without loops:

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

Answer 5

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

Answer 6

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:

.....

Answer 7

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// /#}