Question

我有这个代码...

import java.util.*;

public class SwitchExample {

    public static void main(String[] args) {
        System.out.print("Enter ID: ");
        Scanner scanner = new Scanner(System.in);
        while (!scanner.hasNextInt()) {
            System.out.print("Invalid. Enter again: ");
            scanner.next();
        }

        int number = scanner.nextInt();

        System.out.println("Your number was: " + number);
        System.out.println("Test message got printed!");
    }

}

键入有效输入是可以的，并且键入无效字符后，仍可以按要求运行。但是，在两种情况下键入Enter键都不会引发错误。请帮助我如何实现。我尝试了多种方法，但都没有奏效。

Answer 1

好吧，您只是在扫描整数，然后尝试扫描“下一行”

while(scanner.hasNextLine())

如果您按Enter键，它将捕获，我想是。

Answer 2

要输入，您需要添加一个特殊情况，诸如此类

String line = scanner.nextLine();
if (line .isEmpty()) {
        System.out.println("Enter Key pressed");
}

这是您需要的完整源代码：

public static void main(String[] args) {
    System.out.print("Enter ID: ");
    Scanner scanner = new Scanner(System.in);
    String readString = scanner.nextLine();
    while(readString!=null) {
        System.out.println(readString);

        if (readString.isEmpty()) {
            System.out.println("Read Enter Key.");
        }
        else if (isInteger(readString)) {
            System.out.println("Read integer");
        }
        else{
            System.out.println("Read char");
        }

        if (scanner.hasNextLine()) {
            readString = scanner.nextLine();
        } else {
            readString = null;
        }
    }
}

public static boolean isInteger(String s) {
    Scanner sc = new Scanner(s.trim());
    if(!sc.hasNextInt()) return false;
    // we know it starts with a valid int, now make sure
    // there's nothing left!
    sc.nextInt();
    return !sc.hasNext();
}

在扫描仪验证中将Enter键视为无效

2 个答案: