为什么findInLine（token）有效但hasNext（token）不起作用

Question

I am trying to understand the Scanner class and its methods. I have understood how findInLine method works but I can't understand how hasNext(String) and next(String) methods work.

Here's my code.

import java.util.Scanner;

public class ScannerTest {

    public static void main(String[] args) {

        hasNextTest("\\d");
        //findInLineTest("\\d");
    }

    public static void findInLineTest(String token) {
        Scanner s = new Scanner(System.in);
        System.out.println("Input source string :");
        System.out.flush();
        String str = null;
        while (( str = s.findInLine(token)) != null) {
            System.out.println(str);
        }
        System.out.flush();
        s.close();
    }

    public static void hasNextTest(String token) {
        Scanner s = new Scanner(System.in);
        System.out.println("Input source string :");
        System.out.flush();
        System.out.println(s.hasNext(token));
//      while (s.hasNext(token)) {

//          System.out.println(s.next(token));
//      }
        System.out.flush();
        s.close();
    }

}

The part I can't understand is if I pass to the scanner String "abcd12345abcd" or say any other String that has digits, the method hasNext("\\\\d") doesn't return true. It doesn't return true for any pattern and Strings for which the hasNext(String) method should return true. I'm not using the Scanner object correctly but can someone help me find the mistake.

Thanks.

Answer 1

hasNext(pattern) attempts to match the specified pattern against entire tokens. The string "abcd12345abcd" does not have delimiters, so it constitutes one token. Your pattern looks for a token consisting of a single digit, which obviously isn't going to match.

findInLine(pattern) , on the other hand, attempts to find a pattern anywhere in the line, so the first found digit would match.