在同一行打印

Question

我是 Java 新手，我正在尝试在 1 行上打印学生编号和编号（在本例中为 cijfer）。但由于某种原因，我得到了奇怪的迹象等。另外，当我尝试其他东西时，我得到一个非静态上下文错误。这是什么意思，具体是如何工作的？

下面是我的代码：

import java.text.DecimalFormat;
import java.util.Arrays;


public class Student {
public static final int AANTAL_STUDENTEN = 50;
public  int[] studentNummer = new int[AANTAL_STUDENTEN];
public String[] cijfer;


public int[] StudentNummers() {

    for (int i = 0; i < AANTAL_STUDENTEN; i++) {
        studentNummer[i] = (50060001 + i);

    }
    return studentNummer;
}

 public  String[] cijfers(){

    for (int i = 0; i < AANTAL_STUDENTEN; i++) {
        DecimalFormat df = new DecimalFormat("#.#");

        String cijferformat = df.format(Math.random() * ( 10 - 1 ) + 1);
        cijfer[i++] = cijferformat;
    }
    return cijfer;
}

public static void main(String[] Args) {
    System.out.println("I cant call the cijfer and studentnummer.");
}
}

我也知道我的 cijfer 数组给出了一个空指针异常。我仍然需要解决这个问题。

Answer 1

只需将所有方法和类变量设为static。然后您可以通过main 方法访问它们。此外，您的代码中有一些错误：

public class Student {
    public static final int AANTAL_STUDENTEN = 50;
    // NOTE: static variables can be moved to local ones

    // NOTE: only static method are available from static context
    public static int[] StudentNummers() {
        int[] studentNummer = new int[AANTAL_STUDENTEN];

        for (int i = 0; i < AANTAL_STUDENTEN; i++)
            studentNummer[i] = 50060001 + i;

        return studentNummer;
    }

    // NOTE: only static method are available from static context
    public static String[] cijfers() {
        // NOTE: it is better to use same `df` instance
        DecimalFormat df = new DecimalFormat("#.#");
        String[] cijfer = new String[AANTAL_STUDENTEN];

        for (int i = 0; i < AANTAL_STUDENTEN; i++)
            // NOTE: remove `i++`, because we have it in the loop
            cijfer[i] = df.format(Math.random() * (10 - 1) + 1);

        return cijfer;
    }

    // NOTE: this is `static` method, therefore it has access only to static methods and variables
    public static void main(String[] Args) {
        String[] cijfer = cijfers();
        int[] studentNummer = StudentNummers();

        // TODO you can pring two arrays one element per line
        for(int i = 0; i < AANTAL_STUDENTEN; i++)
            Sytem.out.println(cijfer[i] + '-' + studentNummer[i]);

       // TODO as alternative, you can print whole array
       System.out.println(Arrays.toString(cijfer));
       System.out.println(Arrays.toString(studentNummer));
    }
}

Answer 2

除了我在 cmets 中提到的内容之外，您的设计还需要工作。您有一个 class Student，其中包含 50 个 studentNummer 和 cijfer 成员。据推测，Student 将只有一个studentNummer 和一个cijfer。您需要 2 个类：1 个用于单个 Student，另一个用于保存所有 Student 对象（例如 StudentBody）。

public class StudentBody {
    // An inner class (doesn't have to be)
    public class Student {
        // Just one of these
        public  int studentNummer;
        public String cijfer;

        // A constructor. Pass the student #
        public Student(int id) {
            studentNummer = id;
            DecimalFormat df = new DecimalFormat("#.#");
            cijfer = df.format(Math.random() * ( 10 - 1 ) + 1);
        }

        // Override toString
        @Override
        public String toString() {
            return studentNummer + " " + cijfer;
        }
    }

    public static final int AANTAL_STUDENTEN = 50;
    public Student students[] = new Student[AANTAL_STUDENTEN];

    // StudentBody constructor
    public StudentBody() {
        // Create all Students
        for (int i = 0; i < AANTAL_STUDENTEN; i++) {
            students[i] = new Student(50060001 + i);
        }
    }

    // Function to print all Students
    public void printStudents(){
        for (int i = 0; i < AANTAL_STUDENTEN; i++) {
            System.out.println(students[i]);
       }
    }

    public static void main(String[] Args) {
        // Create a StudentBody object
        StudentBody allStudents = new StudentBody();
        // Print
        allStudents.printStudents();
    }
}

Answer 3

你可以循环System.out.print。否则，使您的函数静态以从 main 访问它们。同时初始化你的 cijfer 数组。

Answer 4

我不是java开发者，但尝试

System.out.print