How to split a string with quotes (like command arguments) in bash?

The simplest solution is using making an array of the quoted args which you could then loop over if you like or pass directly to a command.

eval "array=($string)"

for arg in "${array[@]}"; do echo "$arg"; done   

p.s. Please comment if you find a simpler way without eval.

Edit:

Building on @Hubbitus' answer we have a fully sanitized and properly quoted version. Note: this is overkill and will actually leave additional backslashes in double or single quoted sections preceding most punctuation but is invulnerable to attack.

declare -a "array=($( echo "$string" | sed 's/[][`~!@#$%^&*():;<>.,?/\|{}=+-]/\\&/g' ))"

I leave it to the interested reader to modify as they see fit http://ideone.com/FUTHhj

When I saw David Postill's answer, I thought "there must be a simpler solution". After some experimenting I found the following works:-

string='"aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo"'
echo $string
eval 'for word in '$string'; do echo $word; done'

This works because eval expands the line (removing the quotes and expanding string) before executing the resultant line (which is the in-line answer):

for word in "aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo"; do echo $word; done

An alternative which expands to the same line is:

eval "for word in $string; do echo \$word; done"

Here string is expanded within the double-quotes, but the $ must be escaped so that word in not expanded before the line is executed (in the other form the use of single-quotes has the same effect). The results are:-

[~/]$ string='"aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo"'
[~/]$ echo $string
"aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo"
[~/]$ eval 'for word in '$string'; do echo $word; done'
aString that may haveSpaces IN IT
bar
foo
bamboo
bam boo
[~/]$ eval "for word in $string; do echo \$word; done"
aString that may haveSpaces IN IT
bar
foo
bamboo
bam boo

How do I do that?

$ for l in "aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo"; do echo $l; done
aString that may haveSpaces IN IT
bar
foo
bamboo
bam boo

---

What do I do if my string is in a bash variable?

The simple approach of using the bash string tokenizer will not work, as it splits on every space not just the ones outside quotes:

DavidPostill@Hal /f/test
$ cat ./test.sh
#! /bin/bash
string='"aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo"'
for word in $string; do echo "$word"; done

DavidPostill@Hal /f/test
$ ./test.sh
"aString
that
may
haveSpaces
IN
IT"
bar
foo
"bamboo"
"bam
boo"

To get around this the following shell script (splitstring.sh) shows one approach:

#! /bin/bash 
string=$(cat <<'EOF'
"aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo" 
EOF
)
echo Source String: "$string"
results=()
result=''
inside=''
for (( i=0 ; i<${#string} ; i++ )) ; do
    char=${string:i:1}
    if [[ $inside ]] ; then
        if [[ $char == \\ ]] ; then
            if [[ $inside=='"' && ${string:i+1:1} == '"' ]] ; then
                let i++
                char=$inside
            fi
        elif [[ $char == $inside ]] ; then
            inside=''
        fi
    else
        if [[ $char == ["'"'"'] ]] ; then
            inside=$char
        elif [[ $char == ' ' ]] ; then
            char=''
            results+=("$result")
            result=''
        fi
    fi
    result+=$char
done
if [[ $inside ]] ; then
    echo Error parsing "$result"
    exit 1
fi

echo "Output strings:"
for r in "${results[@]}" ; do
    echo "$r" | sed "s/\"//g"
done

Output:

$ ./splitstring.sh
Source String: "aString that may haveSpaces IN IT" bar foo "bamboo" "bam boo"
Output strings:
aString that may haveSpaces IN IT
bar
foo
bamboo
bam boo

Source: StackOverflow answer Split a string only by spaces that are outside quotes by choroba. Script has been tweaked to match the requirements of the question.