Bash split string on delimiter, assign segments to array

Question

In bash, I would like to transform a PATH-like environment variable that may contain space-separated elements into an array, making sure elements bearing spaces do not cause word-splitting, appearing as "multiple elements".

Let PATH_VARIABLE be the variable in question.

Let un:dodecaedro:per:tirare:per:i danni be the content of the variable.

It is intended for the desired array _to have 6 elements, not 7.

0) un
1) dodecaedro
2) per
3) tirare
4) per
5) i danni

The "tricky" entry may be the space-separated value: i danni.

I am looking for the absolute most elegant and correct way to achieve this.

Limitation: it must work with my bash version: v3.2.48(1)-release

In python this is done just beautifully as so:

>>> v='un:dodecaedro:per:tirare:per:i danni'
>>> len(v.split(':'))
6

Works. Shows what I am looking for.

What's the best way to do this in our beloved bash?

Can you specifically improve on my attempt 4?

Here my attempts

#!/bin/bash

PATH_VARIABLE='un:dodecaedro:per:tirare:per:i danni'

# WRONG
a1=($(echo $PATH_VARIABLE | tr ':' '\n'))

# WRONG
a2=($(
  while read path_component; do
  echo "$path_component"
  done < <(echo "$PATH_VARIABLE" | tr ':' '\n')
))

# WORKS, it is elegant.. but I have no bash 4!
# readarray -t a3 < <(echo "$PATH_VARIABLE" | tr ':' '\n')

# WORKS, but it looks "clunky" to me :(
i=0
while read line; do
  a4[i++]=$line
done < <(echo "$PATH_VARIABLE" | tr ':' '\n')

n=${#a4[@]}
for ((i=0; i < n; i++)); do
  printf '%2d) %s\n' "$i" "${a4[i]}"
done

My environment

bash v3.2.48(1)-release

osx OS X v10.8.3 (build 12D78)

3.2.48-1 — Apple, Mac OS X? My shell reports 3.2.48(1) with parentheses instead of a dash, so maybe not. Why don't the first two options work for you? They both worked for me (and I didn't expect the second to work because I was under the illusion that my bash didn't support process substitution <(...) — maybe the sh version doesn't but the bash version does... yes: sh does not and bash does have process substitution). — Jonathan Leffler
– Jonathan Leffler, Commented Apr 3, 2013 at 3:05
Indeed, Mac OS X "the Cougar". Intellectual slavery. Instrument of subjugation. Anyway, a1 and a2 split to 7 elements, not 6.. right? — Robottinosino
– Robottinosino, Commented Apr 3, 2013 at 3:10
Ah...point out such subtleties so people know why the obvious-seeming answer isn't the one you want. — Jonathan Leffler
– Jonathan Leffler, Commented Apr 3, 2013 at 3:11

kojiro · Accepted Answer · 2013-04-04 11:49:08Z

9

f() {
  local IFS=:
  local foo
  set -f # Disable glob expansion
  foo=( $@ ) # Deliberately unquoted 
  set +f
  printf '%d\n' "${#foo[@]}"
  printf '%s\n' "${foo[@]}"
}

f 'un:dodecaedro:per:tirare:per:i danni'
6
un
dodecaedro
per
tirare
per
i danni

Modifying Jim McNamara's answer, you could just reset IFS:

oIFS="$IFS"
foo='un:dodecaedro:per:tirare:per:i danni'
IFS=: arr=( $foo )
IFS="$oIFS"

I prefer the function scope because it protects IFS changes from bleeding into the global scope without requiring special care to reset it.

Edits and explanations:

As a matter of clarification: In the second example, the IFS setting does change the global variable. The salient difference between this:

IFS=: arr=( $foo )

and this:

IFS=: read -a arr <<< "$foo"

is that the former is two variable assignments and no commands, and the latter is a simple command (see simple command in man (1) bash.)

Demonstration:

$ echo "$BASH_VERSION"
3.2.48(1)-release
$ echo "$IFS"


$ foo='un:dodecaedro:per:tirare:per:i danni'
$ IFS=: read -a arr <<< "$foo"
$ echo "${#arr[@]}"
6
$ echo "$IFS"


$ IFS=: arr1=( $foo )
$ echo "${#arr1[@]}"
6
$ echo "$IFS"
:

edited Apr 4, 2013 at 11:49

answered Apr 3, 2013 at 3:19

kojiro

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12 Comments

Robottinosino Over a year ago

This line works for me: IFS=: foo=( $1 ). 6 elements. No ; before the assignment. Change it to foo=( $1 ) -> 2 elements. Can you try/report back result?

kojiro Over a year ago

IFS=: foo=($1) "works", but it's messy without the function scope, because it spills the value of IFS into the global scope. With the function controlling the scope, however, you don't need to put the things on one line.

Robottinosino Over a year ago

Does not on my systems. Please check and let us know.. (I tried: GNU/Linux (kernel 3.5.0-25-generic) Ubuntu 12.10 bash v4.2.37(1)-release too). IFS does not "spill" beyond the one statement in which it is specified before the assignment..

kojiro Over a year ago

@Robottinosino Does not what? Please clarify what you're asking for.

Robottinosino Over a year ago

I contend that what you wrote could be written more concisely: IFS=: foo=( $1 ). IFS, I say, does not get set globally by prefixing it to the array assignment. I also say there are no other adverse effects of being more "concise" that way. Used within a function or not.

|

ormaaj · Accepted Answer · 2013-04-04 03:06:19Z

7

# Right. Add -d '' if PATH members may contain newlines.
IFS=: read -ra myPath <<<"$PATH"

# Wrong!
IFS=: myPath=($PATH)

# Wrong!
IFS=:
for x in $PATH; do ...

# How to do it wrong right...
# Works around some but not all word split problems
# For portability, some extra wrappers are needed and it's even harder.
function stupidSplit {
    if [[ -z $3 ]]; then
        return 1
    elif [[ $- != *f* ]]; then
        trap 'trap RETURN; set +f' RETURN
        set -f
    fi
    IFS=$3 command eval "${1}=(\$${2})"
}

function main {
    typeset -a myPath
    if ! stupidSplit myPath PATH :; then
        echo "Don't pass stupid stuff to stupidSplit" >&2
        return 1
    fi
}

main

Rule #1: Don't cram a compound data structure into a string or stream unless there's no alternative. PATH is one case where you have to deal with it.

Rule #2: Avoid word / field splitting at all costs. There are almost no legitimate reasons to apply word splitting on the value of a parameter in non-minimalist shells such as Bash. Almost all beginner pitfalls can be avoided by just never word splitting with IFS. Always quote.

edited Apr 4, 2013 at 3:06

answered Apr 3, 2013 at 21:20

ormaaj

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1 Comment

ultracrepidarian Over a year ago

Awesome. Why is this at the bottom? Good thing I kept scrolling... I knew an elegant one-liner exists for this task. Like for most of the tasks in the shell.

jim mcnamara · Accepted Answer · 2013-04-03 03:29:16Z

6

Consider:

$ foo='1:2 3:4 5:6'
$ IFS=':'; arr=($foo)
$ echo "${arr[0]}"
1
$ echo "${arr[1]}"
2 3
$ echo "${arr[2]}"
4 5
$ echo "${arr[3]}"
6

Oh well - took me too long to format an answer... +1 @kojiro.

answered Apr 3, 2013 at 3:29

jim mcnamara

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Bash split string on delimiter, assign segments to array

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