Question

我写了一些代码，我想要一些帮助关于公司的这个程序

这是一个主要的：

Employee e1=new Employee (43208931,"Zainab","Directer","14/5/1990",15000);
    Employee e2=new Employee (43208932,"Ahmed","Maneger","14/5/1991",18000);
    Employee e3=new Employee (43208932,"Talal","Engeniar","14/5/1989",1000);
    Employee e4=new Employee (43208931,"khaled","software Engeniar","14/5/1978",6000);
    Employee e5=new Employee (43208931,"Mohammed","Director","14/5/1978",9000);
    Employee e6=new Employee (43208931,"Jalel","software Designer","14/5/1978",8000);


     Department d1;
     Department d2;
     Department d3;

    d1 = new Department (e1 , e3 , "Jeddah Branch");
    d2 = new Department (e2 , e4 ,"Hail Branch");
    d3 = new Department (e5 , e6 ,"Musqet Branch"); 

     Project p1=new Project (d3,"Bank System");
     Project p2 = new Project (d2,"Employee System");
     Project p3=new Project (d1,"Student System");
     System.out.println(p1.toString());
     System.out.println(p2.toString());
     System.out.println(p3.toString());

这是类Employee：`public class Employee { 私人长身份证; 私有字符串名称; 私人字符串位置; private String DateOfBirth; 私人双薪;

Employee(){

}
public Employee(long ID, String Name, String position, String DateOfBirth, double Salary) {
    this.ID = ID;
    this.Name = Name;
    this.position = position;
    this.DateOfBirth = DateOfBirth;
    this.Salary = Salary;
}

public long getID() {
    return ID;
}

public String getName() {
    return Name;
}

public String getPosition() {
    return position;
}

public String getDateOfBirth() {
    return DateOfBirth;
}

public double getSalary() {
    return Salary;
}

@Override
public String toString() {
    return "Employee{" + "ID=" + ID + ", Name=" + Name + ", position=" + position + ", DateOfBirth=" + DateOfBirth + ", Salary=" + Salary + '}';
}


`

这是班级部门：`公共班级部门扩展员工{ 员工员工; private String DName;

Department(){

}



public Department(Employee Employee,Employee E, String DName) {
    this.Employee = E;
    this.Employee = Employee;
    this.DName = DName;

}

public Department(Employee Employee, Employee E,String DName, long ID, String Name, String position, String DateOfBirth, double Salary) {
    super(ID, Name, position, DateOfBirth, Salary);
    this.Employee = Employee;
    this.DName = DName;

}



 public Employee getEmployee(){
     return Employee;}

 public String getDName(){
     return DName;
 }

@Override
public String toString() {
    return "\n Department:{" + DName + " }" +Employee +  '}';
}

}`

最后这是Class Project：`public class Project extends Department {

Department Department ;
String projecttName;
Project(){

}

public Project(Department Department, String DepartmentName) {
    this.Department = Department;
    this.projecttName = DepartmentName;
}

public Project(Department Department, String DepartmentName, String DName) {
    this.Department = Department;
    this.projecttName = DepartmentName;
}

public Project(Department Department, String DepartmentName, commpany.Employee Employee, String DName, long ID, String Name, String position, String DateOfBirth, double Salary) {
    super(Employee,Employee , DName, ID, Name, position, DateOfBirth, Salary);
    this.Department = Department;
    this.projecttName = DepartmentName;
}

public Department getDepartment() {
    return Department;
}

public String getDepartmentName() {
    return projecttName;
}

@Override
public String toString() {
    return "\n Project :" +projecttName+" {" + Department + '}';
}

} ` 我的问题是：同一部门的第二名员工不是在印刷品＆gt;＆gt; 这个问题是什么？

Answer 1

您将两名员工分配到同一个变量，意味着一个被丢弃

this.Employee = E;
this.Employee = Employee;

如果你想要同样的两个员工，你必须保存它们并在toString（）中打印它们。

提示：如果您想知道如何执行此操作，建议您使用IDE生成此代码。你不必自己写。 ;）

Answer 2

你的问题在这里

public class Department extends Employee { 
Employee Employee; 
private String DName;

    Department(){

    }



    public Department(Employee Employee,Employee E, String DName) {
        this.Employee = E;
        this.Employee = Employee;
        this.DName = DName;

    }

您只有一个实例字段Employee Employee（请注意，变量名称以小写或下划线开头）。

执行此代码时

this.Employee = E;
this.Employee = Employee;

this.Employee换言之，您的私有变量只是被重写

解决方法是使用：两个实例变量Employee e1; Employee e2

然后你的构造函数看起来像

public Department(Employee e1,Employee e2, String DName) {
    this.e1 = e1;
    this.e2 = e2;
    this.DName = DName;

}

此外，如果您想将两名以上的员工添加到特定部门，请使用List<Employee > myList或array

Answer 3

在Department课程中，您只有一个Employee。要拥有多个Employee，您应该使用List。

private List<Employee> employees = new ArrayList<Employee>();

public Department(Employee Employee, Employee E, String DName) {
    this.employees.add(E);
    this.employees.add(Employee);
    this.DName = DName;
}

@Override
public String toString() {
    return "\n Department:{" + DName + " }" + employees + '}';
}

对象第二个员工没有出现

3 个答案: