Java The Complete Reference, Seventh Edition

System.out.println("Modified sequence: " + str);

}

}

The output is shown here:

Original sequence: Jon Jonathan Frank Ken Todd

Modified sequence: Eric Eric Frank Ken Todd

Because the regular expression “Jon.*? “ matches any string that begins with Jon followed

by zero or more characters, ending in a space, it can be used to match and replace both Jon

and Jonathan with the name Eric. Such a substitution is not possible without pattern matching

capabilities.

Using split( )

You can reduce an input sequence into its individual tokens by using thesplit( )method

defined byPattern. One form of thesplit( )method is shown here:

String[ ] split(CharSequencestr)

It processes the input sequence passed instr,reducing it into tokens based on the delimiters

specified by the pattern.

For example, the following program finds tokens that are separated by spaces, commas,

periods, and exclamation points:

// Use split().

import java.util.regex.*;

class RegExpr9 {

public static void main(String args[]) {

// Match lowercase words.

Pattern pat = Pattern.compile("[ ,.!]");

String strs[] = pat.split("one two,alpha9 12!done.");

for(int i=0; i < strs.length; i++)

System.out.println("Next token: " + strs[i]);

}

}

The output is shown here:

Next token: one

Next token: two

Next token: alpha9

Next token: 12

Next token: done

As the output shows, the input sequence is reduced to its individual tokens. Notice that the

delimiters are not included.

832 Part II: The Java Library