Re: More Java Than WebObjects
Re: More Java Than WebObjects
- Subject: Re: More Java Than WebObjects
- From: "Jonathan Fleming" <email@hidden>
- Date: Fri, 11 Jun 2004 17:31:14 +0100
Thanks Michael,
I'm in the process of tesing it, but I think that regex is the way to go, I
had forgotten about that.
Somewhat like you, my regex skills are limited so I'm going to have to do a
bit of digging possibly if your rust has really set in.
Jonathan :^)
From: Michael Engelhart <email@hidden>
To: "Jonathan Fleming" <email@hidden>
CC: email@hidden
Subject: Re: More Java Than WebObjects
Date: Fri, 11 Jun 2004 09:42:28 -0500
Try messing around with java.util.regex.*
Easier to maintain IMO.
Here's a sample that can be tweaked to better suit your needs (i.e. write a
better regex that will do "exactly" what you want). I'm a bit rusty in my
regex skills :-)
import java.util.regex.*;
public class Test {
public static void main(String args[]) {
String matchString = "walls, plumbing, \"painting & decorating\",
decorating, \"concrete mixing truck\", concreting, bricklayers";
System.out.println(matchString);
System.out.println("Start length: " + matchString.length());
Pattern p = Pattern.compile("(\\S+)\\s+");
Matcher m = p.matcher(matchString);
StringBuffer sb = new StringBuffer();
while (m.find()) {
sb.append(m.group().trim());
}
System.out.println(sb);
System.out.println("End length: " + sb.length());
}
}
Mike Engelhart
http://www.eztrip.com/
On Jun 11, 2004, at 6:17 AM, Jonathan Fleming wrote:
What's the best way to write code that will count the spaces between each
word but not prases that are sourounded by double quotation marks eg.
there are 2 phrases in the example below making up a total of 6 words (6
spaces), but how can avoid counting the spaces in the phrases?:
walls, plumbing, "painting & decorating", decorating, "concrete mixing
truck", concreting, bricklayers
This is the code I have so far:
public static String wordCounter(String s, int numOfWords){
int wordCounter=0;
int numToReduce=0;
for (int index=0;index<s.length();index++) {
char c = s.charAt(index);
if (Character.isWhitespace(c)) {wordCounter++;} // Counts the
spaces.
}
if (wordCounter > numOfWords) {
numToReduce=wordCounter - numOfWords;
throw new NSValidation.ValidationException
("You are only allowed to enter "+numOfWords+" words, but you
have entered "+wordCounter+". <br />"+
"You'll need to delete any "+numToReduce+" of these words.");
}
return s;
}
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