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我的删除方法似乎不起作用,因为当用户尝试输入员工姓名时,应用程序无法删除员工。应该发生的情况如下:

  1. 用户使用我编写的名为 userInputByName 的方法输入员工的姓名。
  2. 应用程序在商店中搜索该员工。
  3. 员工被移除。

当我打印出员工仍然在那里的商店时,第 3 步不起作用。

我现在将向您展示我的代码。

MainApp()

//---------------------------------------------------------------------------------------
//  Name:        Case 3: Delete by Name.
//  Description: Choice 3 gives the user an option to delete an employee by name.
//---------------------------------------------------------------------------------------
            case 3:
                System.out.println("Delete by Name.");
                Employee employeeDelete = MenuMethods.userInputByName();
                Store.searchByName(employeeDelete.getEmployeeName());
                System.out.println("Your choice is: "+ employeeDelete);
                Store.remove(employeeDelete);
                break;

员工

//---------------------------------------------------------------------------------------
//  Employee class.
//---------------------------------------------------------------------------------------
public class Employee
{
//---------------------------------------------------------------------------------------
//  Variables to be used in the employee store.
//---------------------------------------------------------------------------------------
    private String employeeName;
    private int employeeId;
    private String employeeEmail;
//---------------------------------------------------------------------------------------
//  Name:        Constructors.
//  Description:
//---------------------------------------------------------------------------------------
    public Employee(String employeeName, int employeeId, String employeeEmail) 
    {
        this.employeeName = employeeName;
        this.employeeId = employeeId;
        this.employeeEmail = employeeEmail;
    }
//---------------------------------------------------------------------------------------
//  Overloading the constructor for the use with userInputByName method.
//---------------------------------------------------------------------------------------
    public Employee(String employeeName) 
    {
        this.employeeName = employeeName;
    }
//---------------------------------------------------------------------------------------
//  Name:   Getters.
//---------------------------------------------------------------------------------------
    public String getEmployeeEmail() 
    {
        return employeeEmail;
    }

    public String getEmployeeName() 
    {
        return employeeName;
    }
    public int getEmployeeId() 
    {
        return employeeId;
    }
//---------------------------------------------------------------------------------------
//  Name:   Setters.
//---------------------------------------------------------------------------------------
    public void setEmployeeEmail(String employeeEmail) 
    {
        this.employeeEmail = employeeEmail;
    }
    public void setEmployeeName(String employeeName) 
    {
        this.employeeName = employeeName;
    }
    public void setEmployeeId(int employeeId)
    {
        this.employeeId = employeeId;
    }

//---------------------------------------------------------------------------------------
//  Name:   toString.
//---------------------------------------------------------------------------------------
    public String toString() 
    {
        return "\t\t\tEmployee\n" +
                "********************************************************************\n"+
                "Employee Name: "+ employeeName +"\n"+ 
                "Employee Id: " + employeeId +"\n"+  
                "Employee Email: " + employeeEmail;
    }
//---------------------------------------------------------------------------------------
}

删除方法

public Employee remove(Employee key) {
        // Remove the Employee by name.
        if (map.containsKey(key))
            return map.remove(key); // if it is there remove and return.
        else
            return null; // if its not there return nothing.
    }

哈希图声明

HashMap<String, Employee> map;
    private static Scanner keyboard = new Scanner(System.in);

    public EmployeeStore() {
        map = new HashMap<String, Employee

按名称搜索

// ---------------------------------------------------------------------------------------
    // Name: Search by Name.
    // //---------------------------------------------------------------------------------------
    public Employee searchByName(String employeeName) {
        Employee employee = map.get(employeeName);
        System.out.println(employee);
        return employee;
    }

用户输入

//---------------------------------------------------------------------------------------
//  Name:        userInputByName.
//  Description: This method is used in the MainApp to give the user capability to search by name.
//---------------------------------------------------------------------------------------
    public static Employee userInputByName() 
    {
        // String temp is for some reason needed. If it is not included
        // The code will not execute properly.
        String temp = keyboard.nextLine();
        Employee e = null;
        System.out.println("Please enter the Employee Name:");
        String employeeName = keyboard.nextLine();

        return e = new Employee(employeeName);

    }

添加到哈希图中

//---------------------------------------------------------------------------------------
//   Create a Store named Store and add Employee's to the Store.
//---------------------------------------------------------------------------------------
        EmployeeStore Store = new EmployeeStore();
        Store.add(new Employee("James O' Carroll", 18, "hotmail.com"));

        Store.add(new Employee("Andy Carroll", 1171, "yahoo.com"));

        Store.add(new Employee("Luis Suarez", 7, "gmail.com"));
4

1 回答 1

2

Hashmap的声明为HashMap<String, Employee> map.

如果要从中删除某些内容,则需要传入 a String,而不是Employee. 尝试类似:

public Employee remove(String key) 
{
    return map.remove(key);
}

HashMap删除之前无需检查是否包含密钥。该方法将为null您返回。

编辑:我很惊讶这并没有为您抛出编译时错误。

编辑2:好的。因此,您正在创建Employee对象,然后将其传递给您的add()方法。这很好,但你需要让你的方法与你所做add()的方法相匹配。remove()因此,如果您正在这样做,EmployeeStore.remove(<employee.getEmployeeName())那么您应该使您的add()方法如下所示:

public Employee add(Employee input)
{
    return map.put(input.getEmployeeName(), input);
}

Employee如果之前在 that 下存储了一个对象,则此函数确实会返回一个对象Key,但您可以选择忽略该值。拥有add()这样的方法应该使其与您的remove()方法相匹配。由于您有一个员工 ID 号,您甚至可以将您的 ID 更改Key为该整数,因为它可能比名称更独特。

于 2012-07-25T14:29:57.087 回答