Questao Scanner

6 de junho de 2007 6 respostas

arthurnn 6 de junho de 2007

ae pessoal outro que eu nao entendi!!!

Determine a saída desse programa:

public class Exerc63 {

public static void main(String[] args) {
	String input = "1 done 2 done all done";
	Scanner s = new Scanner(input);
	s.findInLine("(\\d+) done (\\d+) done (\\w+) done");

	MatchResult result = s.match();

	for (int i=0; i &lt= result.groupCount(); i++) {
		System.out.println(result.group(i));
		s.close();
	}
}

}

6 Respostas

dipeloco 6 de jun. de 2007

1 done 2 done all done
1
2
all

arthurnn 6 de jun. de 2007

sim e agora a explicação?!?!?

Omeganosferatu 8 de jun. de 2007

tbm nao entendi =/

marcosbrandao 8 de jun. de 2007

http://java.sun.com/j2se/1.5.0/docs/api/ :

The scanner can also use delimiters other than whitespace. This example reads several items in from a string:
String input = "1 fish 2 fish red fish blue fish";
     Scanner s = new Scanner(input).useDelimiter("\\s*fish\\s*");
     System.out.println(s.nextInt());
     System.out.println(s.nextInt());
     System.out.println(s.next());
     System.out.println(s.next());
     s.close();
prints the following output:
1
     2
     red
     blue
The same output can be generated with this code, which uses a regular expression to parse all four tokens at once:
String input = "1 fish 2 fish red fish blue fish";
     Scanner s = new Scanner(input);
     s.findInLine("(\\d+) fish (\\d+) fish (\\w+) fish (\\w+)");
     MatchResult result = s.match();
     for (int i=1; i<=result.groupCount(); i++)
         System.out.println(result.group(i);
     s.close();
The default whitespace delimiter used by a scanner is as recognized by Character.isWhitespace.

A scanning operation may block waiting for input.

The next() and hasNext() methods and their primitive-type companion methods (such as nextInt() and hasNextInt()) first skip any input that matches the delimiter pattern, and then attempt to return the next token. Both hasNext and next methods may block waiting for further input. Whether a hasNext method blocks has no connection to whether or not its associated next method will block.

The findInLine(java.lang.String), findWithinHorizon(java.lang.String, int), and skip(java.util.regex.Pattern) methods operate independently of the delimiter pattern. These methods will attempt to match the specified pattern with no regard to delimiters in the input and thus can be used in special circumstances where delimiters are not relevant. These methods may block waiting for more input.

Omeganosferatu 11 de jun. de 2007

Preciso estudar mais regex isso sim =/

Marcos_Henrique_N_Al 11 de jun. de 2007

java.util.Scanner é uma classe pqra facilitar trabalhar com partes de uma string.
por exemplo vc tem:

String variavel = "Marcos^Endereco^Telefone";
Scanner scanner = new Scanner(variavel);

com o scanner criado tem que declarar qual o caracter que separa as partes da variavel, que no caso é “^”, mas pode ser também “|”, “!”, “;”, etc…

scanner.useDelimiter("^");

com os métodos hasNext() e next() que devolve a próxima parte da variavel é possível fazer o seguinte:

while(scanner.hasNext()) {
 System.out.println(scanner.next());
}

Criado 6 de junho de 2007

Ultima resposta 11 de jun. de 2007

Respostas 6

Participantes 5

6 Respostas

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