如何更正我的排序器方法以允许更多用户输入？

Question

我正在尝试为我的几个朋友构建一个排序算法，并且我现在正在为其进行测试构建。最终我会让它变得更加动态，但目前我只是将它接受到用户输入姓名和输入的任何数字的位置。但是，当我运行代码时，它要求输入名称，但随后它会按照输入数字值的指示正确终止。我已经处理这个问题半个小时了，说实话，我有点困惑。

我的代码如下：

//Sorter.java
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Sorter {

    public static void dict() {

        // Create a dictionary
        Map<String, Range> dictionary = new HashMap<>();

        // Add groups and their corresponding value ranges
        dictionary.put("Family A", new Range(0, 10));
        dictionary.put("Family B", new Range(11, 20));
        dictionary.put("Family C", new Range(21, 30));

        // Retrieve the value range for a specific group
        Range rangea = dictionary.get("Family A");
        System.out.println("Value range for Family A: " + rangea.getStart() + " - " + rangea.getEnd());
        Range rangeb = dictionary.get("Family B");
        System.out.println("Value range for Family B: " + rangeb.getStart() + " - " + rangeb.getEnd());
        Range rangec = dictionary.get("Family C");
        System.out.println("Value range for Family C: " + rangec.getStart() + " - " + rangec.getEnd());

    }

    public static String getName() {
        try (Scanner scanner = new Scanner(System.in)) {
            System.out.print("Enter your name: ");
            if (scanner.hasNextLine()) {
                String nameInput = scanner.nextLine();
                System.out.println("Hello, " + nameInput + ".");
                return nameInput;
            } else {
                // handle no input available
                return null;
            }
        }
    }

/**
 * Prompts the user to enter a numerical value and prints the corresponding family group.
 */
    public static void counter() {
        try (Scanner scanner = new Scanner(System.in)) {
            System.out.print("Enter a numerical value: ");
            if (scanner.hasNextLine()) {
                String userInputString = scanner.nextLine();
                try {
                    int userInput = Integer.parseInt(userInputString);
                    String whatFamily = getGroup(userInput);
                    System.out.println("Family: " + whatFamily);
                } catch (NumberFormatException e) {
                    System.out.println("No Family LOL get rekt");
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        }
    }

    /**
     * Determines the group based on the given value.
     *
     * @param  value  the value to be evaluated
     * @return        the group corresponding to the value
     */
    public static String getGroup(int value) {
        // Check if the value is less than 0
        if (value < 0) {
            return "Family A";
        }
        // Check if the value is within the range of Family A
        else if (value >= 1 && value <= 10) {
            return "Family A";
        }
        // Check if the value is within the range of Family B
        else if (value >= 11 && value <= 20) {
            return "Family B";
        }
        // Check if the value is within the range of Family C
        else if (value >= 21 && value <= 30) {
            return "Family C";
        }
        // Check if the value is greater than 30
        else if (value > 30) {
            return "Family C";
        }
        // Value is outside the range of any family
        else {
            return "No Family LOL get rekt";
        }
    }
}

class Range {
    private int start;
    private int end;

    public Range(int start, int end) {
        this.start = start;
        this.end = end;
    }

    public int getStart() {
        return start;
    }

    public int getEnd() {
        return end;
    }
}

//Exec.java

import java.util.HashMap;
import java.util.Map;

public class Exec {
    /**
     * Initializes a dictionary with groups and their corresponding value ranges.
     * Prints the value range for each group and prompts user input.
     *
     * @param  args  command line arguments
     */
    public static void main(String[] args) {

        // Prompt user input
        Sorter.getName();
        Sorter.dict();
        Sorter.counter();
    }
}

Answer 1

您的多个扫描仪导致出现问题。只需使用一个并将其传递给方法即可。

// Prompt user input
Scanner scanner = new Scanner(System.in);
Sorter.getName(scanner);
Sorter.dict();
Sorter.counter(scanner);