I know that if I do print ("f" + 2 * "o") in python the output will be foo.
But how do I do the same thing in a bash script?
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5 Answers
You can use bash command substitution to be more portable across systems than to use a variant specific command.
$ myString=$(printf "%10s");echo ${myString// /m} # echoes 'm' 10 times
mmmmmmmmmm
$ myString=$(printf "%10s");echo ${myString// /rep} # echoes 'rep' 10 times
reprepreprepreprepreprepreprep
Wrapping it up in a more usable shell-function
repeatChar() {
local input="$1"
local count="$2"
printf -v myString '%*s' "$count"
printf '%s\n' "${myString// /$input}"
}
$ repeatChar str 10
strstrstrstrstrstrstrstrstrstr
5 Comments
gniourf_gniourf
Use
printf -v myString '%*s' "$count" instead of myString=$(...). And use local to mark your variables local (or drop the local variables and use the parameters directly).
Inian
@gniourf_gniourf: Yup, it is a pretty old answer. Will make the change anyway!
James T.
You can simplify to
$ ${$(printf %10s)// /rep}Inian
@pjh: Fixed it, turned out to be a wrong edit
pjh
repeatChar works now. Its handling of counts with leading zeros may be surprising to some though. (repeatChar str 09 fails with an invalid octal number error.)You could simply use loop
$ for i in {1..4}; do echo -n 'm'; done
mmmm
Comments
That will do:
printf 'f'; printf 'o%.0s' {1..2}; echo
Look here for explanations on the "multiplying" part.
1 Comment
kvantour
I like this answer most, but I would rewrite it as:
printf -- "f%s\n" $(printf -- "o%.0s" {1..2}) . The downside is that it needs a subshell, but it does the job quite well.In bash you can use simple string indexing in a similar manner
#!/bin/bash
oos="oooooooooooooo"
n=2
printf "%c%s\n" 'f' ${oos:0:n}
output
foo
Another approach simply concatenates characters into a string
#!/bin/bash
n=2
chr=o
str=
for ((i = 0; i < n; i++)); do
str="$str$chr"
done
printf "f%s\n" "$str"
Output
foo
There are several more that can be used as well.
Comments
You can create a function to loop a string for a specific count and use it in the loop you are executing with dynamic length. FYI a different version of oter answers.
line_break()
{
for i in `seq 0 ${count}`
do
echo -n "########################"
done
}
line_break 10
prints: ################
