Java程序分级

Question

我已经在这个程序上工作了几个小时，但我不知道如何让该程序实际打印分数文本文件中的成绩

public class Assign7{
  private double finalScore;
  private double private_quiz1;
  private double private_quiz2;
  private double private_midTerm;
  private double private_final;
  private final char grade;

  public Assign7(double finalScore){
    private_quiz1 = 1.25;
    private_quiz2 = 1.25;
    private_midTerm = 0.25;
    private_final = 0.50;

       if (finalScore >= 90) {
           grade = 'A';
       } else if (finalScore >= 80) {
           grade = 'B';
       } else if (finalScore >= 70) {
           grade = 'C';
       } else if (finalScore>= 60) {
           grade = 'D';
       } else {
           grade = 'F';
       }
}


  public String toString(){
    return finalScore+":"+private_quiz1+":"+private_quiz2+":"+private_midTerm+":"+private_final;

  }
} 

这段代码和这段代码一样可以编译

import java.util.*;
import java.io.*;

public class Assign7Test{
  public static void main(String[] args)throws Exception{

  int q1,q2;
  int m = 0;
  int f = 0;
  int Record ; 
  String name;

    Scanner myIn = new Scanner( new File("scores.txt") );

    System.out.println( myIn.nextLine() +"  avg  "+"letter");

   while( myIn.hasNext() ){
      name = myIn.next();
      q1 = myIn.nextInt();
      q2 = myIn.nextInt();
      m = myIn.nextInt();
      f = myIn.nextInt();
       Record myR = new Record( name, q1,q2,m,f);
       System.out.println(myR);

      }
    }

public static class Record {

       public Record() {
       }

       public Record(String name, int q1, int q2, int m, int f)
       { 

       } 
   } 
}

编译代码后，我得到这个，它不能准确计算我在scores.txt中的数字姓名 quiz1 quiz2 期中期末平均字母

分配7Test$Record@4bcc946b

分配7Test$Record@642423

Exception in thread "main" java.until.InputMismatchException    
    at java.until.Scanner.throwFor(Unknown Source)    
    at java.until.Scanner.next(Unknown Source)    
    at java.until.Scanner.nextInt(Unknown Source)        
    at java.until.Scanner.nextInt(Unknown Source)    
    at Assign7Test.main(Assign7Test.java:25)

Answer 1

抛开异常(exception)，您实际上正在打印Record类型的对象。您需要做的是重写toString()以提供对象的适当表示。

@Override
public String toString() {
    return "Something meaningful about your Record object.";
}

我还注意到，您正在使用 System.out.println('...') 中的 nextLine() 来推进扫描程序。您可能想从代码中注释掉该部分。