java - 创建对象时遇到问题

Question

我试图在我的应用程序的主方法中创建员工类的对象，但它找不到变量部门姓氏名字或薪水，我不确定为什么我的语句看起来像这样

Employee employee = new Employee( department, lastName, firstName, salary);

员工类别看起来像

package javaapplication14;
public class Employee implements DepartmentConstants, Displayable 
{
    private int department;
    private String firstName;
    private String lastName;
    private double salary;

    public Employee(int department, String lastName, String firstName,
        double salary)
    {
        this.department = department;
        this.lastName = lastName;
        this.firstName = firstName;
        this.salary = salary;
    }

    @Override
    public String getDisplayText() 
    {
        String displayText = firstName + lastName + salary + department ;
        return displayText;
    }
}

我不确定是否需要显示其实现的接口(interface)，但如果需要，我将编辑我的帖子以包含它们。主要方法是

package javaapplication14;
public class DisplayableTestApp
{
    public static void main(String args[])
    {
        System.out.println("Welcome to the Displayable Test application\n");

        // create an Employee object
        Employee employee = new Employee( department, lastName, firstName, salary);

        // display the employee information
        String displayTextEmployee = employee.getDisplayText();
        System.out.println(displayTextEmployee);

        // create a Product object
        Product product = new Product();
        // display the product information
        String displayText = product.getDisplayText();
        System.out.println(displayText);
    }
}

Answer 1

I'm trying to create an object of the employee class in the main method of my app but it cant find the variables

它找不到变量，因为它们不存在于构造函数调用的范围内。您必须创建变量并将其传递给构造函数。

Employee 类采用构造函数:

public Employee(int department, String lastName, String firstName, double salary)

举个例子:

Employee employee = new Employee(1, "Smith", "John", 50000);

或者:

int department = 1;
String lastName = "Smith";
String firstName = "John";
double salary = 50000;
Employee employee = new Employee(department, lastName, firstName, salary);