Question

public static int intInput(String prompt, String error) {

    int intInput = 0;  // Variable to be returned

    // Start the scanner
    Scanner keyboard = new Scanner(System.in);

    // This loop check to make sure the user entered positive data
    do {
        System.out.println(prompt);

        // This loop check if the user has entered an int
        while (!keyboard.hasNextInt()) {
            System.out.println(error);
            keyboard.next(); 
        }

        intInput = keyboard.nextInt();

    } while (intInput <= 0);

    // Returns an int
    return intInput;

}

如果用户在不输入任何数据的情况下按Enter键，则不会显示提示。它只会转到下一行，并且不会暗示用户仍可以输入数据。

我在nextLine旁边进行了更改，但仍然收到相同的错误

public static int intInput(String prompt, String error) {

    int intInput = 0;  // Variable to be returned

    // Start the scanner
    Scanner keyboard = new Scanner(System.in);

    // This loop check to make sure the user entered positive data
    do {
        System.out.println(prompt);

        // This loop check if the user has entered an int
        while (!keyboard.hasNextInt()) {
            System.out.println(error);
            keyboard.nextLine(); 
        }

        intInput = keyboard.nextInt();

    } while (intInput <= 0);

    // Returns an int
    return intInput;

}

好的，所以我通读了链接并尝试使用nextLine处理，但是它仍然没有改变

// Method for user int input
public static int intInput(String prompt, String error) {

    int intInput = 0;  // Variable to be returned

    // Start the scanner
    Scanner keyboard = new Scanner(System.in);

    // This loop check to make sure the user entered positive data
    do {
        System.out.println(prompt);

        // This loop check if the user has entered an int
        while (!(keyboard.hasNextInt())) {
            System.out.println(error);
            keyboard.nextLine(); 
        }

        intInput = keyboard.nextInt();
        keyboard.nextLine();
    } while (intInput <= 0);

    // Returns an int
    return intInput;

}

Answer 1

您可以在循环中放入try...catch，如果用户未输入int，则只捕获异常：

public static int intInput(String prompt, String error) {

    int intInput = 0;
    Scanner sc = new Scanner(System.in);
    do {
        try {
            System.out.println("Enter an int");
            intInput = Integer.parseInt(sc.nextLine());
        } catch (Exception e) {
            System.out.println(error);
            intInput = 0;
        }
    } while (intInput == 0);

    return intInput;
}

尽管您可能想进行一些适当的异常处理，但不能捕获通用的Exception，但这更多是一个指导原则。通过在intInput中将0强制为catch，这将意味着它将不会脱离do...while并继续进行直到输入有效的int为止。

Java-用户按下Enter键而不输入int数据时无提示

1 个答案: