Question

I try to split a String into tokens.

The token delimiters are not single characters, some delimiters are included into others (example, & and &&), and I need to have the delimiters returned as token.
StringTokenizer is not able to deal with multiple characters delimiters. I presume it’s possible with String.split, but fail to guess the magical regular expression that will suits my needs.

Any idea ?

Example:

Token delimiters: "&", "&&", "=", "=>", " "  
String to tokenize: a & b&&c=>d  
Expected result: an string array containing "a", " ", "&", " ", "b", "&&", "c", "=>", "d"

— Edit —
Thanks to all for your help, Dasblinkenlight gives me the solution. Here is the “ready to use” code I wrote with his help:

private static String[] wonderfulTokenizer(String string, String[] delimiters) {
  // First, create a regular expression that matches the union of the delimiters
  // Be aware that, in case of delimiters containing others (example && and &),
  // the longer may be before the shorter (&& should be before &) or the regexpr
  // parser will recognize && as two &.
  Arrays.sort(delimiters, new Comparator<String>() {
    @Override
    public int compare(String o1, String o2) {
      return -o1.compareTo(o2);
     }
  });
  // Build a string that will contain the regular expression
  StringBuilder regexpr = new StringBuilder();
  regexpr.append('(');
  for (String delim : delimiters) { // For each delimiter
    if (regexpr.length() != 1) regexpr.append('|'); // Add union separator if needed
    for (int i = 0; i < delim.length(); i++) {
      // Add an escape character if the character is a regexp reserved char
      regexpr.append('\\');
      regexpr.append(delim.charAt(i));
    }
  }
  regexpr.append(')'); // Close the union
  Pattern p = Pattern.compile(regexpr.toString());

  // Now, search for the tokens
  List<String> res = new ArrayList<String>();
  Matcher m = p.matcher(string);
  int pos = 0;
  while (m.find()) { // While there's a delimiter in the string
    if (pos != m.start()) {
      // If there's something between the current and the previous delimiter
      // Add it to the tokens list
      res.add(string.substring(pos, m.start()));
    }
    res.add(m.group()); // add the delimiter
    pos = m.end(); // Remember end of delimiter
  }
  if (pos != string.length()) {
    // If it remains some characters in the string after last delimiter
    // Add this to the token list
    res.add(string.substring(pos));
  }
  // Return the result
  return res.toArray(new String[res.size()]);
}

It could be optimize if you have many strings to tokenize by creating the Pattern only one time.

Answer 1

You can use the Pattern and a simple loop to achieve the results that you are looking for:

List<String> res = new ArrayList<String>();
Pattern p = Pattern.compile("([&]{1,2}|=>?| +)");
String s = "s=a&=>b";
Matcher m = p.matcher(s);
int pos = 0;
while (m.find()) {
    if (pos != m.start()) {
        res.add(s.substring(pos, m.start()));
    }
    res.add(m.group());
    pos = m.end();
}
if (pos != s.length()) {
    res.add(s.substring(pos));
}
for (String t : res) {
    System.out.println("'"+t+"'");
}

This produces the result below:

's'
'='
'a'
'&'
'=>'
'b'