我的删除方法似乎无法正常运行,因为当用户尝试输入员工姓名时,应用程序无法删除该员工。应该发生的事情如下:
当我打印出员工仍在那里的商店时,步骤3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.
}
Hashmap声明
HashMap<String, Employee> map;
private static Scanner keyboard = new Scanner(System.in);
public EmployeeStore() {
map = new HashMap<String, Employee
SearchByName
// ---------------------------------------------------------------------------------------
// Name: Search by Name.
// //---------------------------------------------------------------------------------------
public Employee searchByName(String employeeName) {
Employee employee = map.get(employeeName);
System.out.println(employee);
return employee;
}
UserInput
//---------------------------------------------------------------------------------------
// 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hashmap
//---------------------------------------------------------------------------------------
// 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"));
答案 0 :(得分:2)
您的Hashmap
被声明为HashMap<String, Employee> map
。
如果您想从中移除某些内容,则需要传递String
,而不是Employee
。尝试类似:
public Employee remove(String key)
{
return map.remove(key);
}
在删除密钥之前,无需检查HashMap
是否包含密钥。该方法将为您返回null
。
编辑:我很惊讶这并没有给你带来编译时错误。
edit2:好的。因此,您要创建Employee
对象,然后将其传递到add()
方法中。这样很好,但您需要使用add()
方法匹配您所做的remove()
方法。因此,如果您正在执行EmployeeStore.remove(<employee.getEmployeeName())
,那么您应该使add()
方法看起来像这样:
public Employee add(Employee input)
{
return map.put(input.getEmployeeName(), input);
}
如果之前存储了一个Employee
对象,则此函数会返回Key
对象,但您可以选择忽略该值。使用这样的add()
方法可以使其与remove()
方法匹配。由于您拥有员工ID号,您甚至可以将Key
更改为该整数,因为它可能比名称更独特。