作爲變量傳遞一個類 - 學校示例 - Java

Question

例如，我在繞過傳遞類作爲變量時遇到了困難 - 例如 - 我正在考慮學校以及如何實現一個簡單的系統來跟蹤事物如

• 學生
• 教師
• 主題

使得下列條件：

• 一個學生需要主題
• 教師教一個主題
• 一個主題既有學生和教師

每個類的一個基本的例子是如如下：

Student.java

private String name; 
private int age; 
private Subject subject; 
private String school; // Can implement another class of School later 

public Student(int id, String name, int age, String school, String course) { 
    this.id = id; 
    this.name = name; 
    this.age = age; 
    this.school = school; 
    this.course = subject; 
} 

Teacher.java

private String name; 
private Subject subjectTaught; 
private Date taughtSince; // Date is another thing that perplexes me, but that can be saved for another time 
private int holidaysRemaining; 

public Teacher(int id, String name, Subject subjectTaught, String taughtSince, int holidaysRemaining) { 
    this.id = id; 
    this.name = name; 
    this.subjectTaught = subjectTaught; 
    this.taughtSince = taughtSince; 
    this.holidaysRemaining = holidaysRemaining; 
} 

主題

String name; 
int length; 
Teacher teacher; 
Student student; 

public Subject(int id, String name, int length, Teacher teacher, Student student) { 
    this.id = id; 
    this.name = name; 
    this.length = length; 
    this.teacher = teacher; 
    this.student = student; 
} 

我明白，我可以實現一個ArrayList或主題內的HashMap來跟蹤哪些學生服用哪些科目等。

我只是有點失落，當涉及到實際得到的代碼下來，使事情發生......

的最終結果將是有某種用於CRUD操作的庫存管理系統「，增加學生和教師對象等

任何指導將是偉大的！

Answer 1

我認爲Student.java和Teacher.java應該有Subject[] subjects而不是Subject subject。由於每個學生和老師都與許多主題相關，而不僅僅是一個。

Answer 2

在這種情況下，我不會推薦ArrayList，因爲同一個對象可以在List中多次出現。您可以使用HashSet，在這種情況下，您需要在學生和教師課程中實施hashCode和equals。如果您實施Comparator<Student>和Comparator<Teacher>，則還可以使用TreeSet。假設一個對象是由它的ID字段標識，您也可以使用HashMap：

class Subject { 
    String name; 
    int length; 
    private final Map<Integer,Teacher> teachers = new HashMap<>(); 
    private final Map<Integer,Student> students = new HashMap<>(); 

    public Subject(int id, String name, int length) { 
     this.id = id; 
     this.name = name; 
     this.length = length; 
    } 

    public void addTeacher(Teacher teacher) { 
     teachers.put(teacher.getId(), teacher); 
    } 

    public void removeTeacher(Teacher teacher) { 
     teachers.remove(teacher.getId()); 
    } 

    public Teacher[] getTeachers() { 
     return teachers.values().toArray(new Teacher[teachers.size()]); 
    } 

    // ... same for students 
} 

Answer 3

也許這點你在正確的方向？ 我不確定我是否完全理解你的問題，請讓我知道這是你正在尋找的！

public class School { 

    private String name; 
    private Map<String, Subject> subjects = new HashMap<>(); 

    public School(String name) { 
     this.name = name; 
    } 

    public String getName() { 
     return name; 
    } 

    public void addSubject(Subject subject) { 
     subjects.add(subject.name, subject); 
    } 

    public Subject getSubject(String name) { 
     return subjects.get(name); 
    } 
} 

public class Subject { 

    Map<String, Student> students = new HashMap<>(); 
    String name; 
    int length; 
    Teacher teacher; 

    public Subject(int id, String name, int length, Teacher teacher) { 
     this.id = id; 
     this.name = name; 
     this.length = length; 
     this.teacher = teacher; 
    } 

    public void addStudent(Student student) { 
     students.add(student.name, student); 
    } 

    public void getStudent(String name) { 
     students.get(name); 
    } 
}