I want to write out a bash array variable to a file, with each element on a new line. I could do this with a for loop, but is there another (cleaner) way to join the elements with \n
?
9 Answers
Here's a way that utilizes bash parameter expansion and its IFS
special variable.
$ System=('s1' 's2' 's3' 's4 4 4')
$ ( IFS=$'\n'; echo "${System[*]}" )
We use a subshell to avoid overwriting the value of IFS
in the current environment. In that subshell, we then modify the value of IFS
so that the first character is a newline (using $'...'
quoting). Finally, we use parameter expansion to print the contents of the array as a single word; each element is separated by the first charater of IFS
.
To capture to a variable:
$ var=$( IFS=$'\n'; echo "${System[*]}" )
If your bash is new enough (4.2 or later), you can (and should) still use printf
with the -v
option:
$ printf -v var "%s\n" "${System[@]}"
In either case, you may not want the final newline in var
. To remove it:
$ var=${var%?} # Remove the final character of var
-
-
-
2Shouldn't the last example be
var=${var%?}
instead? This is not a regular expression, so.
will match only a period character.– musiphilSep 13, 2013 at 17:49 -
Need quotes the array variable name:
$ IFS=', ' $ echo ${VAR[*]} first second third $ echo "${VAR[*]}" first,second,third
– SeffJun 16, 2019 at 4:23 -
2
var=$( .. )
trims the trailing newline automatically;printf -v var
doesn't Apr 23, 2020 at 6:09
You can use printf
to print each array item on its own line:
$ System=('s1' 's2' 's3' 's4 4 4')
$ printf "%s\n" "${System[@]}"
s1
s2
s3
s4 4 4
awk -v sep='\n' 'BEGIN{ORS=OFS="";for(i=1;i<ARGC;i++){print ARGV[i],ARGC-i-1?sep:""}}' "${arr[@]}"
or
perl -le 'print join "\n",@ARGV' "${arr[@]}"
or
python -c 'import sys;print "\n".join(sys.argv[1:])' "${arr[@]}"
or
sh -c 'IFS=$'\''\n'\'';echo "$*"' '' "${arr[@]}"
or
lua <(echo 'print(table.concat(arg,"\n"))') "${arr[@]}"
or
tclsh <(echo 'puts [join $argv "\n"]') "${arr[@]}"
or
php -r 'echo implode("\n",array_slice($argv,1));' -- "${arr[@]}"
or
ruby -e 'puts ARGV.join("\n")' "${arr[@]}"
that's all I can remind so far.
Above solutions are pretty much it, but the original question asks for output to file:
$ a=(a b c d e)
$ ( IFS=$'\n'; echo "${a[*]}" ) > /tmp/file
$ cat /tmp/file
a
b
c
d
e
$
Notes: 1) 'echo' provides the final newline 2) If this file will just be read in by bash again, then declare -p may be the serialization wanted.
Using for:
for each in "${alpha[@]}"
do
echo "$each"
done
Using history; note this will fail if your values contain !
:
history -p "${alpha[@]}"
Using basename; note this will fail if your values contain /
:
basename -a "${alpha[@]}"
Using shuf; note that results might not come out in order:
shuf -e "${alpha[@]}"
My take, using only Bash builtins, without mutating IFS
:
# $1 separator
# $2… strings
join_strings () {
declare separator="$1";
declare -a args=("${@:2}");
declare result;
printf -v result '%s' "${args[@]/#/$separator}";
printf '%s' "${result:${#separator}}"
}
Example
$ join_strings $'\n' "a b c" "d e f" "g h i"
a b c
d e f
g h i
You can also use any separator:
$ join_strings '===' "a b c" "d e f" "g h i"
a b c===d e f===g h i
here is my impression of joinArray:
function joinArray() {
local delimiter="${1}"
local output="${2}"
for param in ${@:3}; do
output="${output}${delimiter}${param}"
done
echo "${output}"
}
printf
seems to be the most efficient approach for creating a delimited string from an array:
# create a delimited string; note that printf doesn't put the trailing delimiter
# need to save and restore IFS
# it is prudent to put the whole logic on a single line so as to minimize the risk of future code changes breaking the sequence of saving/restoring of IFS
oldIFS=$IFS; IFS=$'\n'; printf -v var "${arr[*]}"; IFS=$oldIFS
# print string to file; note that the newline is required in the format string because printf wouldn't put a trailing delimiter (which is a good thing)
printf '%s\n' "$var" > file
An even simpler way to do this is:
delim=$'\n'
printf -v var "%s$delim" "${arr[@]}" # create a delimited string
var="${var%$delim}" # remove the trailing delimiter
Example
delim=:
arr=(one two three)
printf -v var "%s$delim" "${arr[@]}" # yields one:two:three:
var="${var%$delim}" # yields one:two_three
-
2
As said before, the following:
$ printf -v var "%s\n" "${System[@]}"
works, if you're converting to a string, but you have to deal with the single blank line at the end. I find you can avoid this by using command substitution instead, so
$ var=$(printf '%\n' "${System[@]}")
should work just fine. Just remember to quote var to get the required effect!