電子郵件方法不起作用

Question

嘿，我有一個EmployeeStore，我已經使用了一個hashmap。地圖存儲的變量是電子郵件名稱和ID。我有一個名爲SearchByEmail的方法，但是這有一個問題。當用戶向用戶界面輸入正確的員工電子郵件時，該方法返回false。

這裏是我的代碼： 這是MainApp

case 2: 
       System.out.println("Search by Email."); 
       Employee employeeSearchEmail = MenuMethods.userInputByEmail(); 
       Store.searchByEmail(employeeSearchEmail.getEmployeeEmail()); 

MenuMethods

//Imports 
import java.util.Scanner; 
//******************************************************************** 

public class MenuMethods 
{ 
    private static Scanner keyboard = new Scanner(System.in); 



    //Methods for the Company Application menu. 
    //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; 
     } 
    //******************************************************************** 
    //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; 
      } 
    //******************************************************************** 
    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); 

    } 
    //******************************************************************** 
    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); 

    } 
    //******************************************************************** 
    public static Employee 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 e = new Employee(employeeEmail); 

    } 
    //******************************************************************** 


} 

SearchByEmail

public boolean searchByEmail(String employeeEmail) 
    { 
      //(for(Employee e : map.values()) {...}) 
      //and check for each employee if his/her email matches the searched value 
      boolean employee = map.equals(employeeEmail);  
      System.out.println(employee); 
      return employee; 

    } 

Answer 1

首先，

​​

確實沒有意義。 map是Hashmap，並且employeeEmail是String。他們在什麼條件下是平等的？

由於您既沒有包含地圖的聲明，也沒有插入新值的代碼，因此您不清楚地圖中存儲的內容以及如何存儲。我現在假設您存儲映射如name -> Employee。如果你想搜索基於電子郵件地址的員工，我建議你做這樣的事情

Employee findByEmail(String email) { 
    for (Employee employee : yourMap.values()) 
     if (employee.getEmail().equals(email)) 
      return employee; 

    // Not found. 
    return null; 
} 

然後檢查是否有email僱員存在，你可以做

public boolean searchByEmail(String employeeEmail) { 
    boolean employee = findByEmail(employeeEmail) != null; 
    System.out.println(employee); 
    return employee; 
} 

Answer 2

我認爲地圖是Map<S,T>對於某些S，T，因此它與employeeEmail不是同一類型，並且具體它不是equals()它。

我懷疑你正在尋找Map.containsValue()（如果電子郵件是在地圖中值）或Map.containsKey()（如果電子郵件是映射的鍵），這取決於究竟map是映射，如果映射到/來自字符串值。

編輯：基於評論澄清：
由於電子郵件是不是在map的關鍵，也不值，建議的解決方案是行不通的，因爲它是。所以你可以選擇其中一個：

1. 使用@ aioobe的解決方案來迭代和檢查每個電子郵件。
2. 向課程添加一個字段：Map<String,Employee> map2它將映射：email_address->employee。鑑於此地圖，您可以使用map2.containsKey(email)搜索電子郵件。它將確保通過電子郵件更快地查找員工，並確保持有額外的地圖。如果我是你，我會選擇這個選擇。