第 28 行的输入代码不起作用

Question

第 28 行的输入代码不起作用，这是怎么回事？当我运行时，它一直运行到第 28 行，然后程序退出。程序编译没有错误。该程序会询问用户的年龄、性别、名字和姓氏，以及他们是否已婚，是否超过 20 岁。如果用户未满 20 岁，程序将不会询问他们是否已婚。

import java.util.Scanner;

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

        //asking the users age, name, and gender 
        System.out.print("What is your gender (M or F): ");
        String gender = Input.nextLine();
        System.out.print("First name: ");
        String FirstName = Input.nextLine();
        System.out.print("Last name: ");
        String LastName = Input.nextLine();
        System.out.print("Age: ");
        int Age = Input.nextInt();

        if (Age >= 20) {  
            System.out.print("Are you married " + FirstName + " (Y or N): ");
            String AreYouMarried = Input.nextLine(); // << PROBLEM

            if (AreYouMarried == "Y") {
                if(gender == "M") {
                    System.out.println("Then I shal call you Mr." + FirstName + " " + LastName + ".");
                }
                else if(gender == "F") {
                    System.out.println("Then I shal call you Ms." + FirstName + " " + LastName + ".");
                }
            }
        }
        if(Age < 20) {
            System.out.print("Then I shall call you " + FirstName + " " + LastName + ".");
        }
    }  
}

Answer 1

1. 这不是在 Java 中比较字符串的正确方法：

   AreYouMarried == "Y"
   

   使用以下内容：

   "Y".equals(AreYouMarried) 
   

   这同样适用于性别比较。

2. 不要在同一个程序中混合调用 nextInt 和 nextLine。
   这会导致更多问题。坚持只使用其中一种
   单个程序中的方法。

Answer 2

发生这种情况的原因是 nextInt() 无法读取按下的最后一个字符，原因是 ENTER = '\n'。 要解决它，只需在询问 mariage 之前调用另一个 nextLine() 即可。会是这样的。。

class Gender {

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

    //asking the users age, name, and gender 
    System.out.print("What is your gender (M or F): ");
    String gender = Input.nextLine();
    System.out.print("First name: ");
    String FirstName = Input.nextLine();
    System.out.print("Last name: ");
    String LastName = Input.nextLine();
    System.out.print("Age: ");
    int Age = Input.nextInt();

    if (Age >= 20) {  
        System.out.print("Are you married " + FirstName + " (Y or N): ");
        Input.nextLine();   
        String AreYouMarried = Input.nextLine(); // << PROBLEM
//HERE CONTINUE CODING

并且不要忘记将“==”更改为 .equals()。 你不能像那样比较字符串。

Answer 3

用equals()方法替换""，比较字符串内容是否相等使用equals()方法

和

在Syso("Are you married " + FirstName + " (Y or N): ");之后添加Input.nextLine();

public static void main(String[] args){

    try {
        Scanner Input = new Scanner(System.in);

        //asking the users age, name, and gender 
        System.out.print("What is your gender (M or F): ");
        String gender = Input.nextLine();
        System.out.print("First name: ");
        String FirstName = Input.nextLine();
        System.out.print("Last name: ");
        String LastName = Input.nextLine();
        System.out.print("Age: ");
        int Age = Input.nextInt();

        if(Age >= 20){  

            System.out.print("Are you married " + FirstName + " (Y or N): ");
            Input.nextLine();
            String AreYouMarried = Input.nextLine(); 

            if(AreYouMarried.equals("Y")){
                if(gender.equals("M") ){
                    System.out.println("Then I shal call you Mr." + FirstName + " " + LastName + ".");
                }else if(gender.equals("F") ){
                    System.out.println("Then I shal call you Ms." + FirstName + " " + LastName + ".");
                }
            }else{
                System.out.println("XYZ msg.");
            }
        }

        if(Age < 20){
            System.out.print("Then I shall call you " + FirstName + " " + LastName + ".");
        }
    } catch (Exception e) {
        // TODO: handle exception
        e.printStackTrace();
    }

}