Bash doesn't parse quotes when converting a string to arguments

Question

This is my problem. In bash 3:

$ test='One "This is two" Three'
$ set -- $test
$ echo $2
"This

How to get bash to understand the quotes and return $2 as This is two and not "This? Unfortunately I cannot alter the construction of the variable called test in this example.

Answer 1

The reason this happens is because of the order in which the shell parses the command line: it parses (and removes) quotes and escapes, then replaces variable values. By the time $test gets replaced with One "This is two" Three, it's too late for the quotes to have their intended effect.

The simple (but dangerous) way to do this is by adding another level of parsing with eval:

$ test='One "This is two" Three'
$ eval "set -- $test"
$ echo "$2"
This is two

(Note that the quotes in the echo command are not necessary, but are a good general practice.)

The reason I say this is dangerous is that it doesn't just go back and reparse for quoted strings, it goes back and reparses everything, maybe including things you didn't want interpreted like command substitutions. Suppose you had set

$ test='One `rm /some/important/file` Three'

...eval will actually run the rm command. So if you can't count on the contents of $test to be "safe", do not use this construct.

BTW, the right way to do this sort of thing is with an array:

$ test=(One "This is two" Three)
$ set -- "${test[@]}"
$ echo "$2"
This is two

Unfortunately, this requires control of how the variable is created.

Answer 2

Now we have bash 4 where it's possible to do something like that:

#!/bin/bash

function qs_parse() { 
    readarray -t "$1" < <( printf "%s" "$2"|xargs -n 1 printf "%s\n" ) 
}

tab='   '  # tabulation here
qs_parse test "One 'This is two' Three -n 'foo${tab}bar'"

printf "%s\n" "${test[0]}"
printf "%s\n" "${test[1]}"
printf "%s\n" "${test[2]}"
printf "%s\n" "${test[3]}"
printf "%s\n" "${test[4]}"

Outputs, as expected:

One
This is two
Three
-n
foo     bar  # tabulation saved

Actually, I am not sure but it's probably possible to do that in older bash like that:

function qs_parse() {
    local i=0
    while IFS='' read -r line || [[ -n "$line" ]]; do
        parsed_str[i]="${line}"
        let i++
    done < <( printf "%s\n" "$1"|xargs -n 1 printf "%s\n" )
}

tab='   ' # tabulation here
qs_parse "One 'This is two' Three -n 'foo${tab}bar'"

printf "%s\n" "${parsed_str[0]}"
printf "%s\n" "${parsed_str[1]}"
printf "%s\n" "${parsed_str[2]}"
printf "%s\n" "${parsed_str[3]}"
printf "%s\n" "${parsed_str[4]}"