Question

我正在尝试创建以下方法来阅读＆＃34; studentmarks.txt＆＃34;文件。但是，我无法将学生的标记作为诸如65 60 52的int读取并存储到数组中。它不断输出错误＆＃34; java.util.InputMismatchException null＆＃34;。如何在不改变＆＃34; studentmarks.txt＆＃34;的格式的情况下解决这个问题。文件？谢谢！

public void readMarksData(String fileName) throws FileNotFoundException
{


    File dataFile = new File(fileName);
    Scanner scanner = new Scanner(dataFile);

    String nameOfCohort = scanner.nextLine(); //1
    System.out.println(nameOfCohort);

    int noOfMarks = scanner.nextInt();  //2
    System.out.println(noOfMarks);

    scanner.nextLine();
    while( scanner.hasNext() )
    {

       scanner.useDelimiter("[,\n]");
       String name = scanner.next();      //3
       System.out.println(name);



         //            int marks[] = new int[3];
         //            for(int i = 0 ; i <= 3 ; i++)
         //            {
         //               marks[i] = scanner.nextInt();
         //            }

       int marks[] = new int[100];

       int markOne = scanner.nextInt();  //4 java.util.InputMismatchException null
       marks = new int[markOne];
       System.out.println(markOne);
       scanner.nextLine();



       int markTwo = scanner.nextInt();      //5
       marks = new int[markTwo];
       scanner.nextLine();

       int markThree = scanner.nextInt();      //6
       marks = new int[markThree];
       scanner.nextLine();
       //              
       //System.out.println(markOne + " " + markTwo + " " + markThree);

    }
    scanner.close();
}

studentmarks.txt：

CS1 Group 2
3
Andreas Antoniades
65 85 77
Charlotte Brocklebank
87 93 81
suzanne dawson
0 55 42

StudentRecord课程：

public class StudentRecord
{

   private String name;
   private String noOfMarks; 
   private int[] marks;

   public StudentRecord(String name)
   {
     marks = new int[24];
     this.name = name;  
   }

Answer 1

int result = Integer.parseInt(number);

您可以使用parseInt(String val)方法将字符串值65解析为整数值，并将数组中的存储解析为

Java扫描程序类将数据文件中的多个int值存储到数组中

1 个答案: