parsing values into array

Question

I have a simple question, I think, but I can't find the solution for it.
A simple perl script prints out following line "tests - Blub" "tests - Blub - Abc" and I assign it to a variable like that var=$(perl ...), but why can't I parse it to an array with varArray=( $var ) and that command varArray=( "tests - Blub" "tests - Blub - Abc" ) works?

The expected result should be like this:

tests - Blub
tests - Blub - Abc

And not like this:

"tests
-
Blub"
"tests
-
Blub
-
Abc"

Thanks for any advices.

glenn jackman · Accepted Answer · 2011-02-18 21:01:56Z

2

Here's some doodling:

$ bash
$ a='"tests - Blub" "tests - Blub - Abc"'
$ ary=( $a ); echo ${#ary[@]}
8
$ ary=( "$a" ); echo ${#ary[@]}
1
$ eval ary=( $a ); echo ${#ary[@]}
2

Clearly the 3rd result is what you want. When you populate your variable from the output of the Perl script, the double quotes within it have no special meaning to the shell: they're just characters. You have to get the shell to parse it (with eval) so that their meaning is exposed.

answered Feb 18, 2011 at 21:01

glenn jackman

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user3458 · Accepted Answer · 2011-02-18 20:04:24Z

1

I did the following:

$ cat >/tmp/test.sh
echo '"tests - Blub" "tests - Blub - Abc"'

$ chmod +x /tmp/test.sh

$ /tmp/test.sh
"tests - Blub" "tests - Blub - Abc"

$ a=`/tmp/test.sh`

$ echo $a
"tests - Blub" "tests - Blub - Abc"

$ arr=($a)

$ echo $arr[1]
"tests[1]

This tells me that () construct ignores double quotes after variable expansions. Moreover, when I do

for i in $a; do echo $i; done

I get a similar result:

"tests
-
Blub"
"tests
-
Blub
-
Abc"

Looks like quotes are handled before variable substitution takes place and not looked at again later.

answered Feb 18, 2011 at 20:04

user3458

4 Comments

CSchulz Over a year ago

That is my problem I am facing to and I don't know how to fix it, because I don't want to switch the language.

user3458 Over a year ago

I hate to say this, but using arrays in Bash is probably a sign you need a more capable language...

CSchulz Over a year ago

There is a solution with eval see the other answer.

user3458 Over a year ago

Ouch. Be careful with $, ` and other characters meaningful inside "".

tooly · Accepted Answer · 2011-02-19 13:38:29Z

0

How about using xargs in combination with sh -c '...' ?

line='"tests - Blub" "tests - Blub - Abc"'
printf '%s' "$line" | xargs sh -c 'printf "%s\n" "$@"' argv0
IFS=$'\n'
ary=( $(printf '%s' "$line" | xargs sh -c 'printf "%s\n" "$@"' argv0) )
echo ${#ary[@]}
printf '%s\n' "${ary[@]}"

answered Feb 19, 2011 at 13:38

tooly

1

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