Question

我试图将扫描仪设置为字符串eventType，以便它习惯于用户的输入，但是我在编辑器上始终收到错误消息。我添加了第一个if语句，以了解我要执行的操作。

import java.util.Scanner;
public class main {   

    public static void main (String[] args) {
        int whatToEat;
        int partySize;
        String eventType;

        Scanner sc = new Scanner(System.in);
        String eventType = sc.next();

        Scanner scnr = new Scanner(System.in);
        whatToEat = scnr.nextInt();

        Scanner scnr1 = new Scanner(System.in);
        partySize = scnr1.nextInt();



        if (eventType.equals("Casual") && partySize == 1) {

            System.out.println("Since you're hosting a casual event for "
                    + partySize +
                    "participants, you should serve sandwiches prepared in the mircowave. ");

Answer 1

如果我进行了更改，那么此代码对我来说很好用。您将两次定义String变量eventType。如果将行String eventType = sc.next();更改为eventType = sc.next();（去掉单词'String'），此代码似乎可以正常工作。

请注意，给您的班级main命名不是一个好主意，但是在这种情况下，这似乎没问题。

您无需为每个输入创建一个新的Scanner对象。这是您的代码版本，可解决我提到的小问题：

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String eventType = sc.next();
        int whatToEat = sc.nextInt();
        int partySize = sc.nextInt();

        if (eventType.equals("Casual") && partySize == 1) {
            System.out.println("Since you're hosting a casual event for "
                    + partySize +
                    " participants, you should serve sandwiches prepared in the mircowave. ");
        }
    }
}

结果：

Casual
1
1
Since you're hosting a casual event for 1 participants, you should serve sandwiches prepared in the mircowave.

如何将扫描仪设置为字符串

1 个答案: