该程序同时要求学生姓名和年级

Question

我正在编写一个程序，要求6名学生的姓名和成绩，并计算出它（A，B，C，F等）。

我有一个恼人的bug，我无法解决。 如果您运行它并查看输出结果，您将知道我的意思：它会跳过询问第二个学生的姓名。

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

尝试使用kb.next()而不是kb.nextLine()。这应该工作。

当您在键入int值后按Enter键，因为它不是一个整数，所以它会被用作kb.nextLine()的值，因此它被跳过。

Answer 2

您的输入包含整数和行。 nextInt会给你提供的下一个int，如果需要的话，消费一个换行符。但是，如果您按照nextInt和newLine执行操作，则newline将满足用于输入int的回车符。

解决的办法是阅读为每个学生的最终成绩后添加

kb.nextLine(); 

。

Answer 3

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(); 
...