我正在建立一个员工店。 employeeStore中的删除功能不起作用。
,而不是删除员工,它不会做任何事情这是我的代码: 主
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;
删除方法
// ---------------------------------------------------------------------------------------
// Name: Remove.
// ---------------------------------------------------------------------------------------
public Employee remove(Employee key) {
// Remove the Employee by name.
if (map.containsKey(key))
return map.remove(key); // if its there remove and return
else
return null; // if its not there return 0x00000000 address
}
菜单方法
//---------------------------------------------------------------------------------------
// Name: Imports.
// Description: To allow the use of different Java classes.
//---------------------------------------------------------------------------------------
import java.util.Scanner;
//---------------------------------------------------------------------------------------
public class MenuMethods
{
private static Scanner keyboard = new Scanner(System.in);
//---------------------------------------------------------------------------------------
// Methods for the Company Application menu.
//---------------------------------------------------------------------------------------
//---------------------------------------------------------------------------------------
// Name: getMenuChoice.
// Description: Method for validating the choice.
//---------------------------------------------------------------------------------------
public static int getMenuChoice(String menuString, int limit,String prompt, String errorMessage)
{
System.out.println(menuString);
int choice = inputAndValidateInt(1, limit, prompt, errorMessage);
return choice;
}
//---------------------------------------------------------------------------------------
// Name: inputAndValidateInt.
// Description: This method is used in the getMenuChoice method.
//---------------------------------------------------------------------------------------
public static int inputAndValidateInt(int min, int max, String prompt,String errorMessage)
{
int number;
boolean valid;
do
{
System.out.print(prompt);
number = keyboard.nextInt();
valid = number <= max && number >= min;
if (!valid)
{
System.out.println(errorMessage);
}
} while (!valid);
return number;
}
//---------------------------------------------------------------------------------------
// Name: userInput
// Description: This method is used in the MainApp to give the user capability to enter
// the details when adding details of an employee into the store.
//---------------------------------------------------------------------------------------
public static Employee userInput()
{
String temp = keyboard.nextLine();
Employee e = null;
System.out.println("Please enter the Employee Name:");
String employeeName = keyboard.nextLine();
System.out.println("Please enter the Employee ID:");
int employeeId = keyboard.nextInt();
temp = keyboard.nextLine();
System.out.println("Please enter the Employee E-mail address:");
String employeeEmail = keyboard.nextLine();
return e = new Employee(employeeName, employeeId, employeeEmail);
}
//---------------------------------------------------------------------------------------
// 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);
}
//---------------------------------------------------------------------------------------
// Name: userInputByEmail
// Description: This method is used in the MainApp to give the user capability to search by email.
//---------------------------------------------------------------------------------------
public static String userInputByEmail()
{
// 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 Email:");
String employeeEmail = keyboard.nextLine();
// This can use the employeeName's constructor because java accepts the
// parameters instead
// of the name's.
return employeeEmail;
}
//---------------------------------------------------------------------------------------
}
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;
}
//---------------------------------------------------------------------------------------
}
答案 0 :(得分:3)
此方法
public static Employee userInputByName
正在创建新 Employee
。 (它还将它无意义地分配给局部变量......代码样式在这里非常奇怪 - 就像对nextLine()
的无关调用一样。)
除非您在equals
中覆盖了hashCode
和Employee
,否则它不会认为 new Employee
对象等于HashMap
中现有的一个。遗憾的是,您还没有向我们展示Employee
类,但是如果它们具有相同的名称,则可能是为了将员工进行比较(并给出相同的哈希码)。
答案 1 :(得分:1)
您需要做的就是
public Employee remove(Employee key) {
// Remove the Employee by name.
return map.remove(key); // if its there remove and return
}
此问题的最常见原因是hashCode和equals不正确,不匹配或依赖于可变字段。
答案 2 :(得分:1)
经典:你可能在Employee
对象上缺少equals()和hashcode()。您还应该考虑为地图使用一个简单的密钥,例如employee.id:就像现在一样,您的EmployeeMap可以更好地建模为EmployeeSet。
在代码中进一步查看,您可能希望创建多个Maps作为员工对象的索引:
Map<Integer, Employee> employeeById
Map<String, Employee> employeeByEmail;
MultiMap<String, Employee> employeeByName;
请注意最后一行中建议的com.google.common.collect.Multimap,以便可能包含多个共享相同John Doe名称的员工参考。