如何獲得我的學生成績計算器程序中類似成績字母的總數

Question

我的程序計算30位學生的成績（六個不同的分數），然后顯示每位學生的字母成績。 我已經這樣做了，但是我的問題是如何計算A，B，C，D和F的數量。

我希望我的輸出是這樣的：

A的數目為：

B的數目為：

C的數目為：

D的數目為：

F的數量為：

import javax.swing.*;
import java.awt.*;
import java.io.*;
import java.util.Scanner;

public class FinalGrade extends JPanel
{
        public static String firstName[]= new String[30];
        public static String lastName[] = new String[30];
        public static String grade[]=new String[30];
        public static int HW1;
        public static int HW2;
        public static int HW3;
        public static int Project;
        public static int Midterm;
        public static int Final;
        public static double Avg_homework;
        public static double Avg_exam;
        public static double Final_numeric_grade;


     public FinalGrade()
     {
     super(new GridLayout(1,0));

     String[] columnNames = {"First Name", "Last Name", "Final Grade"};

     Object[][] data = {
     {firstName[0], lastName[0], grade[0]},
     {firstName[1], lastName[1], grade[1]},
     {firstName[2], lastName[2], grade[2]},
     {firstName[3], lastName[3], grade[3]},
     {firstName[4], lastName[4], grade[4]},
     {firstName[5], lastName[5], grade[5]},
     {firstName[6], lastName[6], grade[6]},
     {firstName[7], lastName[7], grade[7]},
     {firstName[8], lastName[8], grade[8]},
     {firstName[9], lastName[9], grade[9]},
     {firstName[10], lastName[10], grade[10]},
     {firstName[11], lastName[11], grade[11]},
     {firstName[12], lastName[12], grade[12]},
     {firstName[13], lastName[13], grade[13]},
     {firstName[14], lastName[14], grade[14]},
     {firstName[15], lastName[15], grade[15]},
     {firstName[16], lastName[16], grade[16]},
     {firstName[17], lastName[17], grade[17]},
     {firstName[18], lastName[18], grade[18]},
     {firstName[19], lastName[19], grade[19]},
     {firstName[20], lastName[20], grade[20]},
     {firstName[21], lastName[21], grade[21]},
     {firstName[22], lastName[22], grade[22]},
     {firstName[23], lastName[23], grade[23]},
     {firstName[24], lastName[24], grade[24]},
     {firstName[25], lastName[25], grade[25]},
     {firstName[26], lastName[26], grade[26]},
     {firstName[27], lastName[27], grade[27]},
     {firstName[28], lastName[28], grade[28]},
     {firstName[29], lastName[29], grade[29]}
      };

     final JTable table = new JTable(data, columnNames);

     table.setPreferredScrollableViewportSize(new Dimension(300,400));

     table.setFillsViewportHeight(true);

     JScrollPane scrollPane = new JScrollPane(table);
     add(scrollPane);
     }

      private static void createAndShowGUI() {
        //Create and set up the window.
        JFrame frame = new JFrame("Grade Display");
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        //Create and set up the content pane.
        FinalGrade newContentPane = new FinalGrade();
        newContentPane.setOpaque(true); //content panes must be opaque
        frame.setContentPane(newContentPane);

        //Display the window.
        frame.pack();
        frame.setVisible(true);
    }

    public static void main(String[] args) throws FileNotFoundException
    {
        Scanner input = new Scanner (new File ("student_grades_input.txt"));

        int count = 0;
        while (input.hasNext())
        {
            firstName[count] = input.next();
            lastName[count] =  input.next();
            HW1 = input.nextInt();
            HW2 = input.nextInt();
            HW3 = input. nextInt();
            Project = input.nextInt();
            Midterm = input.nextInt();
            Final = input.nextInt();


            Avg_homework = (HW1 + HW2 + HW3)/3;

            Avg_exam = (Midterm + Final)/2;


            Final_numeric_grade = 0.45 * Avg_homework + 0.25 * Project + 0.30 * Avg_exam ;

            if (Final_numeric_grade > 89)

                  grade[count] = "A";


           else if (Final_numeric_grade > 79 && Final_numeric_grade < 90)

                  grade[count] = "B";



          else if (Final_numeric_grade > 69 && Final_numeric_grade < 80)

                  grade[count] = "C";


          else if (Final_numeric_grade > 59 && Final_numeric_grade < 70)

             grade[count] = "D";


            else

             grade[count] = "F";


            count++;
            }

        input.close();
        createAndShowGUI();                


    }

}

Answer 1

您可以使用數組存儲成績的計數，其中第一個索引為“ A”，第二個索引為“ B”。 等等。您甚至可以在設置grade數組時執行此操作。 這樣的事情應該可以解決問題：

int[] gradeCount = new int[5]; //Define before your while (input.hasNext())
...
if (Final_numeric_grade > 89){
    grade[count] = "A";
    gradeCount[0]++;}
else if (Final_numeric_grade > 79 && Final_numeric_grade < 90){
    grade[count] = "B";
    gradeCount[1]++;}
else if (Final_numeric_grade > 69 && Final_numeric_grade < 80){
    grade[count] = "C";
    gradeCount[2]++;}
else if (Final_numeric_grade > 59 && Final_numeric_grade < 70){
    grade[count] = "D";
    gradeCount[3]++;}
else{
    grade[count] = "F";
    gradeCount[4]++;}

然后，當您需要訪問成績計數時，請執行以下操作：

System.out.println("Number of A: " + gradeCount[0]);
System.out.println("Number of B: " + gradeCount[1]);
System.out.println("Number of C: " + gradeCount[2]);
System.out.println("Number of D: " + gradeCount[3]);
System.out.println("Number of F: " + gradeCount[4]);

Answer 2

您要做的就是在設置完所有成績並擁有6個計數器（每個字母1個）之后遍歷grade數組：

int numberOfA = 0;
int numberOfB = 0;
int numberOfC = 0;
int numberOfD = 0;
int numberOfE = 0;
int numberOfF = 0;
for (int i = 0; i < grade.length; i++) {
    if (grade[i].equals("A")) {
        numberOfA++;
    } else if (grade[i].equals("B")) {
        numberOfB++;
    } else if (grade[i].equals("C")) {
        numberOfC++;
    } else if (grade[i].equals("D")) {
        numberOfD++;
    } else if (grade[i].equals("E")) {
        numberOfE++;
    } else {
        numberOfF++;
    }
}

然后只需打印或執行您需要對計數器做的任何事情：

System.out.println("Number of A: " + numberOfA);
System.out.println("Number of B: " + numberOfB);
System.out.println("Number of C: " + numberOfC);
System.out.println("Number of D: " + numberOfD);
System.out.println("Number of E: " + numberOfE);
System.out.println("Number of F: " + numberOfF);

還請遵循Java命名約定

除變量外，所有實例，類和類常量均以小寫首字母混合使用。 內部單詞以大寫字母開頭。 變量名稱不應以下划線_或美元符號$字符開頭，即使兩者都允許。

從上Variable names should start with lower case while classes names should start with an upper case ： Variable names should start with lower case while classes names should start with an upper case

---

編輯

就像@Gonzo 所說的那樣，您可以在已有的代碼上使用計數器，這將提高性能，因為您不會在同一數組上進行兩次迭代，您可以這樣做：

if (Final_numeric_grade > 89) { //Be sure to add curly brackets
    grade[count] = "A";
    numberOfA++;
} //And close them
else if (Final_numeric_grade > 79 && Final_numeric_grade < 90) {
    grade[count] = "B";
    numberOfB++;
} 
//And so on ...

並按我上面所述打印它們。 只需確保在增加變量之前初始化變量即可：）

Answer 3

有很多方法可以做到這一點。 一種簡單的方法是為每個成績定義計數器變量，並在確定成績時在while循環中增加它們。 就像是 ：

int gradeACounter=0, gradeBCounter=0;

 if (Final_numeric_grade > 89){

              grade[count] = "A";
              gradeACounter++;
 }

       else if (Final_numeric_grade > 79 && Final_numeric_grade < 90){

              grade[count] = "B";
              gradeBCounter++;
}