获取错误消息“线程“主”java.util.InputMismatchException 中的异常”

Question

public static void main(String[] args) {    Scanner sc = new Scanner(System.in);    int iteam = sc.nextInt();    Student[] students = new Student[iteam];    int id;    String name;    String city, s_city;    double marks, s_marks;    for (int i = 0; i < iteam; i++) {        id = sc.nextInt();        name = sc.nextLine();        city = sc.nextLine();        marks = sc.nextDouble();        students[i] = new Student(id, name, city, marks);    }}我有一个错误“Exception in thread "main" java.util.InputMismatchException    at java.util.Scanner.throwFor(Unknown Source)    at java.util.Scanner.next(Unknown Source)    at java.util.Scanner.nextDouble(Unknown Source)"我也尝试使用 next() 但存在相同的错误。我还导入了 java.util.Scanner。

Answer 1

这里没有很多信息，但我看到一个问题肯定会导致您的问题。

当您从一行中读取一个数值，并期望用户点击返回来提交该值时，您必须先消耗行尾才能继续。扫描仪不会自动执行此操作。要让用户输入 4 个值，在每次输入后按回车键，您的输入循环应如下所示：

for (int i = 0; i < iteam; i++) {

    id = sc.nextInt(); sc.nextLine();

    name = sc.nextLine();

    city = sc.nextLine();

    marks = sc.nextDouble(); sc.nextLine();

    students[i] = new Student(id, name, city, marks);

}

Answer 2

当您读取 4 个值时，您需要sc.nextLine()在从输入中读取第一个 int 后调用。 Scanner#nextInt()只读取整数，不读取换行符"\n"；您将需要使用Scanner#nextLine.

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);

    int iteam = sc.nextInt();

    Student[] students = new Student[iteam];

    int id;

    String name;

    String city, s_city;

    double marks, s_marks;

    for (int i = 0; i < iteam; i++) {

        id = sc.nextInt();

        sc.nextLine();//call next line to prevent exception

        name = sc.nextLine();

        city = sc.nextLine();

        marks = sc.nextDouble();

        students[i] = new Student(id, name, city, marks);

    }

}

完整的工作示例：

import java.util.Scanner;

public class MyClass {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int iteam = sc.nextInt();

        Student[] students = new Student[iteam];

        int id;

        String name;

        String city, s_city;

        double marks, s_marks;

        for (int i = 0; i < iteam; i++) {

            id = sc.nextInt();

            sc.nextLine();

            name = sc.nextLine();

            city = sc.nextLine();

            marks = sc.nextDouble();

            students[i] = new Student(id, name, city, marks);

        }

        for(Student s: students){

            System.out.println(s);

        }

    }

    private static class Student{

        private int id;

        private String name;

        private String city;

        private double marks;

        public Student(int id, String name, String city, double marks){

            this.id = id;

            this.name = name;

            this.city = city;

            this.marks = marks;

        }

        public String toString(){

            return "[Student] Id: "+this.id+", Name: "+this.name+", City: "+this.city+", Mark: "+this.marks;

        }

    }

}

Answer 3

以前的解决方案并不完全正确。经过一点调试会话后，我想出了以下解决方案：

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);

    int iteam;

    iteam = sc.nextInt();

    Student[] students = new Student[iteam];

    int id;

    String name;

    String city;

    double marks;

    for (int i = 0; i < iteam; i++) {

        id = sc.nextInt();

        name = sc.next();

        city = sc.next();

        marks = sc.nextDouble();

        students[i] = new Student(id, name, city, marks);

    }

}

给你一点解释：方法next()，nextInt()并nextDouble()扫描下一个标记作为输入值。但最重要的是 Return 不算数，因为您的 bash 将其解释为您的输入。

此外，您的键盘输入是本地化解释的，特别是如果您想读取双精度。例如，英文分隔符写为DOT，而德文则写为COMMA。