String split()

[Reshma Reddy]

Hi All,
I want to split string into diff parts

  String name = "welcome@ to$ java world!";  

Expected out put is: welcome
to
java
world

I want to eleminate all special character in my string, where as in split we can mention any one character to check in string. how can I do this.

[Bear Bibeault]

What did you find when you perused the String javadoc?

[Reshma Reddy]

bear,
I couldn't get any thing about my requirement, give me any clue to slove this

[Nishan Patel]

Hi Vardhan,

use regex pattern.

Make one pattern which contains your special character. If your string contains this character split it.

May be Regix pattern will help you...

[Rob Spoor]

Why specify your own special characters? There are special character classes that represent all actual letters or punctuation characters; check out the API of java.util.regex.Pattern for more info.

[Mukesh Ranjan]

StringTokenizer will help you.
Below is the code:
String name = "welcome@ to$ java world!";
StringTokenizer tokenizer = new StringTokenizer(name,"@ $");
while(tokenizer.hasMoreTokens()){
System.out.println(tokenizer.nextToken());
}

[zahid zubair]

Try this:

        String name = "welcome@ to$ java world!";
        String[] requiredString = name.split("[^a-zA-Z][ ]*");
        
        byte stringLength = (byte) requiredString.length;
        for (byte i = 0; i < stringLength; i++) {
            System.out.println(requiredString[i]);
        }

[Reshma Reddy]

thanks zhid,

it is working perfect..

[Rob Spoor]

I'd prefer the character class [\s\W]. \s means whitespace ([ \t\n\x0B\f\r]), whereas \W is a synonym for [^\w] and \w is a synonym for [a-zA-Z_0-9]. Of course you can't use * since then all empty strings between all characters will match as well; the regex will be [\s\W]+. And because \W implies \s, \W+ will suffice.

Of course, once you start using other word characters like é you will run into problems, since that also matches \W.