Question

我正在尝试编写一个程序，要求6名学生的姓名和成绩，并计算它（A，B，C，F等..）

我有一个我无法解决的恼人的错误。如果你运行它并看到输出你会知道我的意思：它跳过要求第二个学生名字。

import java.util.*;

public class grading {
    public static void main (String [] args) {
        Scanner kb = new Scanner (System.in);
        String a, b, c, d, e, f = "";
        int mid1, mid2, finl, total = 0;
        int mid12, mid22, finl2, total2 = 0;
        int mid13, mid23, finl3, total3 = 0;
        int mid14, mid24, finl4, total4 = 0;
        int mid15, mid25, finl5, total5 = 0;
        int mid16, mid26, finl6, total6 = 0;
        String g1, g2, g3, g4, g5, g6 = "0";

        System.out.println ("Enter you first student name :");
        a = kb.nextLine();

        System.out.println ("His first midterm grade [out of 25] :");
        mid1 = kb.nextInt();

        System.out.println ("His second midterm grade [out of 25] :");
        mid2 = kb.nextInt();

        System.out.println ("His final test grade [out of 50] :");
        finl = kb.nextInt();

        total = mid1 + mid2 + finl;

        ///////////////////////// Student 1

        System.out.println ("Enter you second student name :");
        b = kb.nextLine();

        System.out.println ("His first midterm grade [out of 25] :");
        mid12 = kb.nextInt();

        System.out.println ("His second midterm grade [out of 25] :");
        mid22 = kb.nextInt();

        System.out.println ("His final test grade [out of 50] :");
        finl2 = kb.nextInt();

        total2 = mid12 + mid22 + finl2;

        ////////////////////// Student 2

        System.out.println ("Enter you third student name :");
        c = kb.nextLine();

        System.out.println ("His first midterm grade [out of 25] :");
        mid13 = kb.nextInt();

        System.out.println ("His second midterm grade [out of 25] :");
        mid23 = kb.nextInt();

        System.out.println ("His final test grade [out of 50] :");
        finl3 = kb.nextInt();

        total3 = mid13 + mid23 + finl3;

        ///////////////////// Student 3

        System.out.println ("Enter you fourth student name :");
        d = kb.nextLine();

        System.out.println ("His first midterm grade [out of 25] :");
        mid14 = kb.nextInt();

        System.out.println ("His second midterm grade [out of 25] :");
        mid24 = kb.nextInt();

        System.out.println ("His final test grade [out of 50] :");
        finl4 = kb.nextInt();

        total4 = mid14 + mid24 + finl4;

        //////////////////////// Student 4

        System.out.println ("Enter you fifth student name :");
        e = kb.nextLine();

        System.out.println ("His first midterm grade [out of 25] :");
        mid15 = kb.nextInt();

        System.out.println ("His second midterm grade [out of 25] :");
        mid25 = kb.nextInt();

        System.out.println ("His final test grade [out of 50] :");
        finl5 = kb.nextInt();

        total5 = mid15 + mid25 + finl5;

        /////////////////// Student 5


        System.out.println ("Enter you sixth student name :");
        f = kb.nextLine();

        System.out.println ("His first midterm grade [out of 25] :");
        mid16 = kb.nextInt();

        System.out.println ("His second midterm grade [out of 25] :");
        mid26 = kb.nextInt();

        System.out.println ("His final test grade [out of 50] :");
        finl6 = kb.nextInt();

        total6 = mid16 + mid26 + finl6;

        /////////////// Student 6 
    }
}

Answer 1

您的输入包含整数和行。 nextInt将为您提供下一个int，如果需要，会使用换行符。但是，如果您使用nextInt关注newLine，则newline会对用于输入int的回车信息感到满意。

解决方案是添加

kb.nextLine();

阅读每个学生的最终成绩后。

Answer 2

另一种解决方案是在每次阅读时使用新的扫描仪。

a = new Scanner (System.in).nextLine();
mid1 = new Scanner (System.in).nextInt();
mid2 = new Scanner (System.in).nextInt();
finl = new Scanner (System.in).nextInt();
b = new Scanner (System.in).nextLine();
...

Answer 3

尝试使用kb.next()代替kb.nextLine()。这应该有用。

在输入int值后按Enter键，因为它不是整数，它会被用作kb.nextLine()的值，因此会被跳过。

该计划同时要求学生姓名和成绩

3 个答案: