Converting a Bash array into a delimited string
Solution 1:
Because parentheses are used to delimit an array, not a string:
ids="1 2 3 4";echo ${ids// /|}
1|2|3|4
Some samples: Populating $ids
with two strings: a b
and c d
ids=("a b" "c d")
echo ${ids[*]// /|}
a|b c|d
IFS='|';echo "${ids[*]}";IFS=$' \t\n'
a b|c d
... and finally:
IFS='|';echo "${ids[*]// /|}";IFS=$' \t\n'
a|b|c|d
Where array is assembled, separated by 1st char of $IFS
, but with space replaced by |
in each element of array.
When you do:
id="${ids[@]}"
you transfer the string build from the merging of the array ids
by a space to a new variable of type string.
Note: when "${ids[@]}"
give a space-separated string, "${ids[*]}"
(with a star *
instead of the at sign @
) will render a string separated by the first character of $IFS
.
what man bash
says:
man -Len -Pcol\ -b bash | sed -ne '/^ *IFS /{N;N;p;q}'
IFS The Internal Field Separator that is used for word splitting
after expansion and to split lines into words with the read
builtin command. The default value is ``<space><tab><newline>''.
Playing with $IFS
:
declare -p IFS
declare -- IFS="
"
printf "%q\n" "$IFS"
$' \t\n'
Literally a space
, a tabulation
and (meaning or) a line-feed
. So, while the first character is a space. the use of *
will do the same as @
.
But:
{
IFS=: read -a array < <(echo root:x:0:0:root:/root:/bin/bash)
echo 1 "${array[@]}"
echo 2 "${array[*]}"
OIFS="$IFS" IFS=:
echo 3 "${array[@]}"
echo 4 "${array[*]}"
IFS="$OIFS"
}
1 root x 0 0 root /root /bin/bash
2 root x 0 0 root /root /bin/bash
3 root x 0 0 root /root /bin/bash
4 root:x:0:0:root:/root:/bin/bash
Note: The line IFS=: read -a array < <(...)
will use :
as separator, without setting $IFS
permanently. This is because output line #2
present spaces as separators.
Solution 2:
You can use printf
too, without any external commands or the need to manipulate IFS:
ids=(1 2 3 4) # create array
printf -v ids_d '|%s' "${ids[@]}" # yields "|1|2|3|4"
ids_d=${ids_d:1} # remove the leading '|'