How do I change a bash array into a string with this pattern? Let's say I have arr=(a b c d) and I want to change to this pattern
'a' + 'b' + 'c' + 'd' with the white space in between.
PS-I figured out this pattern 'a'+'b'+'c'+'d' but not sure how to put " + " instead of just "+" in between.
In pure bash without resorting to an external command and without creating a subshell, using bash built-in printf's implicit loop with -v option (which assigns the output to the variable rather than printing it):
printf -v str " + '%s'" "${arr[@]}"
str=${str:3} # to strip off leading ' + '
echo "$str"
This will also work with array elements containing blank characters (try with arr=(a b c d "x y") ). This solution assumes arr is not an empty array.
Like this:
#!/bin/bash
arr=(a b c d)
str="${arr[@]}"
str=${str// / + } # Parameter Expansion
sed "s/[a-z]/'&'/g" <<< "$str"
'a' + 'b' + 'c' + 'd'
See: http://mywiki.wooledge.org/BashFAQ/073 and "Parameter Expansion" in man bash. Also see http://wiki.bash-hackers.org/syntax/pe.
Here’s a version in pure Bash that also tolerates whitespace in arr members.
arr=(a b c d 'e f' 'g h')
aux=("${arr[@]/#/ \'}") # add prefix " '"
aux=("${aux[@]/%/\' }") # add suffix "' "
str="$(IFS=+ # join with "+"
printf '%s' "${aux[*]}")"
str="${str:1:${#str} - 2}" # trim spaces
printf '%s\n' "$str"
