将单词从字符串拆分为数组，但如果在斜线之间则不拆分

Question

I have this code:

String path; 
path = main.getInput(); // lets say getInput() is "Hello \Wo rld\"
args = path.split("\\s+");

for (int i = 0; i < args.length; i++) {
     System.out.println(args[i]);
}

Is there a way to split the string so that the words are split and put into an array, but only if they are not in between two backslashes, so that "Wo rld" will be one word and not two?

Answer 1

You could try splitting only on spaces that are followed by an even number of backslashes. Raw regex:

\s+(?=(?:[^\\]*\\[^\\]*\\)*[^\\]*$)

Java escaped regex:

\\s+(?=(?:[^\\\\]*\\\\[^\\\\]*\\\\)*[^\\\\]*$)

ideone demo

Answer 2

Try this one:

String s = "John Hello \\Wo rld\\ our world";
Pattern pattern = Pattern.compile("(\\\\.*?\\\\)|(\\S+)");
Matcher m = pattern.matcher(s);
while (m.find( )) {
    if(m.group(1) != null){
        System.out.println(m.group(1));
    }
    else{
        System.out.println(m.group(2));
    }
}

Output:

John
Hello
\Wo rld\
our
world

Answer 3

If it doesn't have to be regex then you can use this simple parser and get your result in one iteration .

public static List<String> spaceSplit(String str) {
    List<String> tokens = new ArrayList<>();

    StringBuilder sb = new StringBuilder();
    boolean insideEscaped = false; //flag to check if I can split on space 

    for (char ch : str.toCharArray()) {

        if (ch == '\\') 
            insideEscaped = !insideEscaped;

        // we need to split only on spaces which are not in "escaped" area
        if (ch == ' ' && !insideEscaped) {
            if (sb.length() > 0) {
                tokens.add(sb.toString());
                sb.delete(0, sb.length());
            }
        } else //and add characters that are not spaces from between \
            sb.append(ch);
    }
    if (sb.length() > 0)
        tokens.add(sb.toString());

    return tokens;
}

Usage:

for (String s : spaceSplit("hello \\wo rld\\"))
    System.out.println(s);

Output:

hello
\wo rld\