java - 扫描仪跳过输入，可能有空格？

Question

这个问题已经有答案了:

Scanner is skipping nextLine() after using next() or nextFoo()? (25 个回答)

已关闭 7 年前。

我刚刚开始使用 Java，想要修改语法。每当我在gender中输入“F”时和age大于或等于 20 应该提示我输入用户是否已婚，由于某种原因扫描仪会跳过它。其他一切都正常。

我得到的输出:

Whats is your gender (M or F): F
First name: Kim
Last name: Kardashian
Age: 32

Are you married, Kim (Y or N)? 
Then I shall call you Ms. Kardashian.

代码:

import java.util.Scanner;

public class GenderGame 
{

    public static void main(String[] args) 
    {
        Scanner sc = new Scanner(System.in);

        int age = 0;
        String Gender = null, fName = null, lName = null, M = null, type = null;

        System.out.print("Whats is your gender (M or F): ");
        Gender = sc.nextLine();
        Gender = Gender.toUpperCase();

        System.out.print("First name: ");
        fName = sc.nextLine();

        System.out.print("Last name: ");
        lName = sc.nextLine();

        System.out.print("Age: ");
        age = sc.nextInt();

        if(Gender.equals("F") && age >= 20)
        {
            System.out.print("\nAre you married, " + fName + " (Y or N)? ");
            M = sc.nextLine();
            M = M.toUpperCase();

            if(M.equals("Y"))
            {
                type = "Mrs. ";
                type = type.concat(lName);
            }
            else
            {
                type = "Ms. ";
                type = type.concat(lName);
            }
        }
        else if(Gender.equals("F") && age < 20)
        {
            type = fName.concat(" " + lName);
        }
        else if(Gender.equals("M") && age >= 20)
        {
            type = "Mr. ";
            type = type.concat(lName);
        }
        else if(Gender.equals("M") && age < 20)
        {
            type = fName.concat(" " + lName);
        }
        else
        {
            System.out.println("There was incorrect input. EXITING PROGRAM");
            System.exit(1);
        }

        System.out.println("\nThen I shall call you " +type+ ".");

    }

}

Answer 1

Scanner 的 nextInt() 方法将换行符保留下来，也就是说，它不会消耗它。这个换行符由您的 nextLine() 方法使用，这就是为什么您看不到它等待您的输入。

要避免这种情况，请在 age = sc.nextInt(); 之后调用 sc.nextLine()，然后保持其余代码不变。

        ...
        System.out.print("Age: ");
        age = sc.nextInt();
        sc.nextLine(); //added

        if(Gender.equals("F") && age >= 20)
        {
            System.out.print("\nAre you married, " + fName + " (Y or N)? ");
            M = sc.nextLine();
            M = M.toUpperCase();

            if(M.equals("Y"))
            {
                type = "Mrs. ";
                type = type.concat(lName);
            }
         ...