searchByEmail方法不起作用

Question

我有這個方法的問題。當用戶請求時，它不輸出正確的搜索。

這裏是我的代碼：

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

public Employee searchByEmail(String employeeEmail) { 
    for (Employee employee : map.values()) { 
     System.out.println(employee); 
     map.equals(getClass()); 
     map.equals(employee.getEmployeeEmail()); 
     employee = new Employee(employeeEmail); 
     ; 
     return employee; 
    } 
    return null; 
} 

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); 

} 

Answer 1

的問題是，有沒有條件，如果在你的程序是這樣的：

public Employee searchByEmail(String employeeEmail) { 
     for (Employee employee : map.values()) { 
      map.equals(getClass()); 
      if (map.equals(employee.getEmployeeEmail())){ 
       System.out.println(employee); 
       return employee; 
      } 
     } 
     return null; 
    } 

這一行： 的System.out.println （僱員）;

，直到找到了比賽，當比賽將返回該員工的對象..

Answer 2

你應該這樣說

if(employeeEmail.equals(employee.getEmployeeEmail()) return employee; 

沒有必要創建Employee對象的新實例。

Answer 3

您想要返回具有特定電子郵件地址的員工。

public Employee searchByEmail(String employeeEmail) 
    { 
      for(Employee employee : map.values()) 
      { 
       if(employee.getEmployeeEmail().equalsIgnoreCase(employeeEmail.trim())) 
          return employee; 
      } 
      return null; 
    } 

順便說一句，什麼是地圖的鍵：因此，如果當前員工的電子郵件地址，等於給定的電子郵件地址，你應該只返回。如果電子郵件地址的關鍵是，你可以簡單地返回：

return map.get(employeeEmail);