Question

我已经在线查看，但我无法解决我遇到的问题。我正在使用hashMap创建一个员工商店，但我不确定我是否应该打印toString（）或theEntry.getKey（），就像我在当前版本中一样。当使用getKey（）方法时，我得到错误的输出：

* 公司员工。 *** 员工姓名：James O'Carroll 员工ID：James O'Carroll 电子邮件：James O'Carroll

正如您在MainApp中看到的，我想要Store.add（新员工（“James O'Carroll”，18，“hotmail.com”））;成为输出。

我会粘贴我的代码：

//MainApp.
public class MainApp
{

    public static void main(String[] args)
    {
        new MainApp().start();

    }
    public void start()
    {
        EmployeeStore Store = new EmployeeStore();
        Store.add(new Employee ("James O' Carroll", 18,"hotmail.com"));
        Store.print();

    }

}

//Employee
//Imports:

//********************************************************************
//Employee Class.
public class Employee
{
//Variables.
    private String employeeName;
    private int employeeId;
    private String employeeEmail;
//********************************************************************  
//Constructor.
    public Employee(String employeeName, int employeeId, String employeeEmail) 
    {
        this.employeeName = employeeName;
        this.employeeId = employeeId;
        this.employeeEmail = employeeEmail;
    }
//********************************************************************
//Getters.
    public String getEmployeeEmail() {
        return employeeEmail;
    }
    public void setEmployeeEmail(String employeeEmail) {
        this.employeeEmail = employeeEmail;
    }
    public String getEmployeeName() {
        return employeeName;
    }
    public int getEmployeeId() {
        return employeeId;
    }
//********************************************************************
//toString method.
    public String toString() {
        return "Employee [employeeName=" + employeeName + ", employeeId="
                + employeeId + ", employeeEmail=" + employeeEmail + "]";
    }
//********************************************************************





}

//EmployeeStore.
//Imports.
import java.util.HashMap;
//********************************************************************
import java.util.Map;

public class EmployeeStore 
{
    HashMap<String, Employee> map;

//Constructor.  
    public EmployeeStore()
    {
        map = new HashMap<String,Employee>();
    }
//********************************************************************
//Hashmap Methods.
//Add to the Hashmap : Employee.
    public void add(Employee obj)
    {

        map.put(obj.getEmployeeName(), obj);
    }
//********************************************************************
//Remove from the Hashmap : Employee.
    public void remove(String key)
    {
      //Remove the Employee by name.
        map.remove(key);
    }
//********************************************************************
//Print the Hashmap : Employee. 
    public void print()
    {
        System.out.println("\n********Employee's in the Company.********");
        for(Map.Entry<String, Employee> theEntry : map.entrySet())
        {
            System.out.println("Employee Name:\t" + theEntry.getKey());
            System.out.println("Employee Id:\t" + theEntry.getKey());
            System.out.println("E-mail:\t "+ theEntry.getKey());

        }
    }


//********************************************************************  
//********************************************************************


}

Answer 1

这显然是错误的：

for(Map.Entry<String, Employee> theEntry : map.entrySet())
{
    System.out.println("Employee Name:\t" + theEntry.getKey());
    System.out.println("Employee Id:\t" + theEntry.getKey());
    System.out.println("E-mail:\t "+ theEntry.getKey());
}

您打印三次相同的值 - 如何是名称，ID 和电子邮件地址？您正在从条目中打印键，该名称始终是名称。您需要条目的值。

你想要这个：

// This assumes you've renamed all the properties in Employee to be more
// sensible - there's no point in including the "employee" part - we already
// know we're calling a method on Employee
for (Employee employee : map.values())
{
    System.out.println("Employee Name:\t" + employee.getName());
    System.out.println("Employee Id:\t" + employee.getId());
    System.out.println("E-mail:\t "+ employee.getEmail());
}

（或者只是打印employee当然。这取决于你想要输出的内容。）

Answer 2

您需要迭代地图的值。通过这样做，您将获得Employee个实例作为回报，并且可以简单地定义您的print()方法，如下所示，

public void print(){
    for(Employee e : map.values()){
        System.out.println(e);
    }
}

请注意，您无需明确调用toString()，因为println会自动为您执行此操作。

HashMapStore。打印方法未显示正确的输出

2 个答案: