Question

我想知道是否有人可以解释为什么我的代码中的情况2和3似乎没有做任何事情，并且可能提供一些建议。我会提供我认为相关的任何信息，如果您要求更多详细信息，我会添加更多信息。

在发布我的代码之前，让我提供一些细节：我的程序旨在成为一个非常简单的员工数据库。它使用switch语句在命令行上提供用户选项。交换机的选项是：

将员工添加到数据库
列出数据库中的所有员工，查看工资和工作小时数
列出所有员工并显示公司利益（如果有）
终止程序

案例1将Employee对象添加到Employee类型的ArrayList中。 Employee类负责跟踪员工姓名，工资，工时和他们工作的公司。

案例1和案例4似乎正常运作。

然而，案例2和3似乎没有做任何事情。以下是整个交换机，它位于包含main方法的驱动程序类中：

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

    int number = 0;

    while(number != 4)
    {

        System.out.print("Please select an option: " +
                "\n1) Add an Employee" +
                "\n2) List Employees" +
                "\n3) List Benefit Status" +
                "\n4) Quit"+ "\n");
        number = keyboard.nextInt();
        keyboard.nextLine();

        switch (number)
        {
            case 1:
                System.out.println("Hourly, contract, or salary employee? ");
                type = keyboard.nextLine();

                if(type.equalsIgnoreCase("hourly"))
                {
                    System.out.print("\nEnter the company: ");
                    comp = keyboard.nextLine();
                    System.out.print("\nEnter the first name: ");
                    fn = keyboard.nextLine();
                    System.out.print("\nEnter the last name: ");
                    ln = keyboard.nextLine();
                    System.out.print("\nEnter the hourly wage: ");
                    wage = keyboard.nextDouble();
                    System.out.print("\nEnter the hours worked: ");
                    hours = keyboard.nextInt();

                    Employee employee2 = new Employee(comp, fn, ln);
                    HourlyEmployee he = new HourlyEmployee(wage, hours);
                }

                else if(type.equalsIgnoreCase("contract"))
                {
                    System.out.print("\nEnter the company: ");
                                        comp = keyboard.nextLine();
                                        System.out.print("\nEnter the first name: ");
                                        fn = keyboard.nextLine();
                                        System.out.print("\nEnter the last name: ");
                                        ln = keyboard.nextLine();
                                        System.out.print("\nEnter the hourly wage: ");
                                        wage = keyboard.nextDouble();
                                        System.out.print("\nEnter the hours worked: ");
                                        hours = keyboard.nextInt();

                                        Employee employee2 = new Employee(comp, fn, ln);
                    ContractEmployee ce = new ContractEmployee(wage, hours);
                }

                else if (type.equalsIgnoreCase("salary"))
                {
                                        System.out.print("\nEnter the company: ");
                                        comp = keyboard.nextLine();
                                        System.out.print("\nEnter the first name: ");
                                        fn = keyboard.nextLine();
                                        System.out.print("\nEnter the last name: ");
                                        ln = keyboard.nextLine();
                                        System.out.print("\nEnter the salary: ");
                    salary = keyboard.nextDouble();

                    Employee employee2 = new Employee(comp, fn, ln);
                    SalaryEmployee se = new SalaryEmployee(salary);
                }

                else
                {
                    System.out.println("Invalid input.");
                    System.exit(0);
                }

                break;
            case 2:
                for(int i = 0; i < employees.size(); i++)
                {
                    System.out.println(employees.get(i).toString());
                }
                break;
            case 3:
                for(int i = 0; i < employees.size(); i++)
                {
                    System.out.println(employees.get(i).determineBenefits());
                }
                break;
            case 4:
                System.exit(0);
                break;
            default:
                System.out.println("Invalid input.");
                System.exit(0);
        }
    }
}

}

在第2和第3种情况下，我试图将ArrayList的索引分别作为参数传递给toString（）方法和determineBenefits（）方法。当这些方法与交换机分开测试时，它们相同才能正常工作。这是toString（）方法：

    public String toString()
{
    return firstName + " " + lastName + " from " + company +
            ". The worker's pay this week was $" + pay + ".";
}

还有determineBenefits（）方法：

    public String determineBenefits()
{
    String benefits;

    if(isSalaryEmployee == true)
    {
        benefits = "This employee has a standard company health " +
            "insurance policy.";
    }

    else if(hhh  >= 40)
    {
        benefits = "This worker gets benefits.";
    }

    else
    {
        benefits = "No benefits.";
    }

    return benefits;
}

和Employee构造函数，如果相关：

public Employee()
{

}

public Employee(String com, String first, String last)
{
    setCompany(com);
    setFirstName(first);
    setLastName(last);
}

那么，我该如何传递位于ArrayList中的Employee对象？

Answer 1

正如评论中指出的，我的问题是我忘了将Employee对象实际添加到ArrayList，这是一个非常简单的修复错误。感谢所有回答的人。

Java - 开关：案例2和3不起作用

1 个答案: