Converting a Bash array into a delimited string

问题

I would like to know the following;

Why the given non-working example doesn't work.
If there are any other cleaner methods than those given in working example.

Non-working example

> ids=(1 2 3 4);echo ${ids[*]// /|}
1 2 3 4
> ids=(1 2 3 4);echo ${${ids[*]}// /|}
-bash: ${${ids[*]}// /|}: bad substitution
> ids=(1 2 3 4);echo ${"${ids[*]}"// /|}
-bash: ${"${ids[*]}"// /|}: bad substitution

Working example

> ids=(1 2 3 4);id="${ids[@]}";echo ${id// /|}
1|2|3|4
> ids=(1 2 3 4); lst=$( IFS='|'; echo "${ids[*]}" ); echo $lst
1|2|3|4

In context, the delimited string to be used in a sed command for further parsing.

回答1:

# REVISION: 2017-03-14
# Use of read and other bash specific features (bashisms)

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:

set | grep ^IFS=
IFS=$' \t\n'

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:

    # OIFS="$IFS"
    # IFS=$': \t\n'
    # unset array 
    # declare -a array=($(echo root:x:0:0:root:/root:/bin/bash))

    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.

回答2:

Your first question is already addressed in F. Hauri's answer. Here's canonical way to join the elements of an array:

ids=( 1 2 3 4 )
IFS=\| eval 'lst="${ids[*]}"'

Some people will cry out loud that eval is evil, yet it's perfectly safe here, thanks to the single quotes. This only has advantages: there are no subshells, IFS is not globally modified, it will not trim trailing newlines, and it's very simple.

回答3:

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 '|'

回答4:

An utility function to join arguments array into a delimited string:

#!/usr/bin/env bash

# Join arguments with delimiter
# @Params
# $1: The delimiter string
# ${@:2}: The arguments to join
# @Output
# >&1: The arguments separated by the delimiter string
array::join() {
  (($#)) || return 1 # At least delimiter required
  local -- delim="$1" str IFS=
  shift
  str="${*/#/$delim}" # Expand arguments with prefixed delimiter (Empty IFS)
  echo "${str:${#delim}}" # Echo without first delimiter
}

declare -a my_array=( 'Paris' 'Berlin' 'London' 'Brussel' 'Madrid' 'Oslo' )

array::join ', ' "${my_array[@]}"
array::join '*' {1..9} | bc # 1*2*3*4*5*6*7*8*9=362880 Factorial 9

declare -a null_array=()

array::join '== Ultimate separator of nothing ==' "${null_array[@]}"

Output:

Paris, Berlin, London, Brussel, Madrid, Oslo
362880

来源：https://stackoverflow.com/questions/13470413/converting-a-bash-array-into-a-delimited-string

标签

arrays

string

bash

delimited-text