int p = s.replace('\n', ' ').indexOf (' ');

Matt Humphrey http://www.iviz.com/

On Mon, 28 Jan 2008 19:38:10 GMT, Roedy Green
<see_website@mindprod.com.invalid> wrote, quoted or indirectly quoted
someone who said :

I handled it by adding a method to the StringTools class.

   /**
    * find the first instance of whitespace (space, \n, \r, \t in a
string.
    *
    * @param s           string to scan
    * @param startOffset where in string to start looking
    *
    * @return -1 if not found, offset relative to start of string
where found, not relative to startOffset.
    */
   public static int indexOfWhiteSpace( String s, int startOffset )
       {
       final int length = s.length();
       for ( int i = startOffset; i < length; i++ )
           {
           switch ( s.charAt( i ) )
               {
               case ' ':
               case '\n':
               case '\r':
               case '\t':
                   return i;
               default:
                   // keep looking
               }
           }// end for
       return -1;
       }

import java.util.*;
import java.util.regex.*;

public class test3 {
    public static void main(String[] args) {
        String str = "nowisthetimeforall good";

        System.out.println(indexOfRegex(str,"\\s+"));
    }

    public static int indexOfRegex(String str, String regex) {
        int retcod = -1;

        Pattern p = Pattern.compile(regex);
        Matcher m = p.matcher(str);
        if (m.find())
            retcod = m.start();

        return retcod;
    }
}