Split a string by whitespace, keeping quoted segments, allowing escaped quotes

You can change your regex to:

keywords = keywords.match(/\w+|"(?:\\"|[^"])+"/g);

Instead of [^"]+ you've got (?:\\"|[^"])+ which allows \" or other character, but not an unescaped quote.

One important note is that if you want the string to include a literal slash, it should be:

keywords = 'pop rock "hard rock" "\\"dream\\" pop"'; //note the escaped slashes.

Also, there's a slight inconsistency between \w+ and [^"]+ - for example, it will match the word "ab*d", but not ab*d (without quotes). Consider using [^"\s]+ instead, that will match non-spaces.

ES6 solution supporting:

• Split by space except for inside quotes
• Removing quotes but not for backslash escaped quotes
• Escaped quote become quote
• Can put quotes anywhere

Code:

keywords.match(/\\?.|^$/g).reduce((p, c) => {
        if(c === '"'){
            p.quote ^= 1;
        }else if(!p.quote && c === ' '){
            p.a.push('');
        }else{
            p.a[p.a.length-1] += c.replace(/\\(.)/,"$1");
        }
        return  p;
    }, {a: ['']}).a

Output:

[ 'pop', 'rock', 'hard rock', '"dream" pop' ]