【问题标题】:How to get Field Variable on Java Spring如何在 Java Spring 上获取字段变量
【发布时间】:2020-08-15 16:40:47
【问题描述】:

我的学习项目有一个问题,例如“如果条件值为空,然后如果条件值字段为空”,例如我的代码遵循以下代码:

对于实体Users.java

@Entity
public class Users {
    private Long id;
    private String employeeId;
    private String fullName;
    private String username;
    private String password;
    ...

    public Users() {
    }

    Some Code Setter and Getter....
}

对于实体Employee.java

@Entity
public Class Employee {
    private Long id;
    private String employeeId;
    private String fullName;
    ...
    
    public Employee() {
    }
    
    Some Code Setter and Getter....
}

然后对于我的类服务,我有使用存储库插入数据员工的案例。如果我们在将数据插入表 Employee 之前有验证数据,我们需要检查 table users not null 然后在字段 employeeId 上应该为 null。我的代码如下:

对于存储库 UserRepo.javaEmployeeRepo.java

@Repository
public interface EmployeeRepo extends CrudRepository<Employee, Long> {

}

@Repository
public interdace UsersRepo extends CrudRepository<Users, Long> {

@Transactional
@Modifying(clearAutomatically = true, flushAutomatically = true)
@Query("UPDATE Users u SET u.employeeId = :employeeId WHERE u.id = :id")
public void updateEmployeeIdUsers(@Param("id") Long id, @Param("employeeId") String employeeId);

}

对于服务UsersService.java

@Service("usersService")
public class UsersService {
    
    @Autowired
    private UsersRepo repo;
    
    public Optional<Users> findById(Long id) {
        return repo.findById(id);
    }
    
    public void updateEmployeeIdUsers(Long id, String employeeId) {
        repo.updateEmployeeIdUsers(id, employeeId);
    }

}

对于服务EmployeeService.java

@Service("employeeService")
public class EmployeeService {
    
    @Autowired
    private EmployeeRepo employeeRepo;
    
    @Autowired
    private UsersService userService;
    
    public Employee insertEmployee(Employee employee) throws Exception {
        Optional<Users> users = userService.findById(employee.getId());
        Users userOptional = new Users(); **//on this my problem**
        userOptional.getEmployeeId(); **//on this my problem**
        if (!users.isPresent()) {
            throw new Exception("User ID : "+ employee.getId() +" Not Founded");
        }else if (!(userOptional == null)) { **//on this my problem**
            throw new Exception("User employeID : "+ employee.getEmployeeId() +" Already Exist on Users");
        }
        
        String str1 = "TEST";
        Long idUser = employee.getId();
        userService.updateEmployeeIdUsers(idUser, str1);
        return employeeRepo.save(employee);
    }

}

在此代码上,如果 userOptional 始终为 NULL,则我们在 else 上有问题,我尝试调试以查看 employeeId 的值,只是我看到的始终为 Null。所以对我的问题有任何想法,因为我尝试某些案例总是因我的问题而失败。如果对我的问题有任何想法,请回复这些我的问题。非常感谢您回答我的问题。

【问题讨论】:

  • userOptional optional 为空,因为此对象为空。 Users userOptional = new Users(); 在这种情况下,您只需创建空对象。我不明白你的代码的目标是什么。
  • 你的代码总是会失败,因为(!(userOptional == null)) 总是true。您创建了一个非空实例Users userOptional = new Users()
  • @Seldo97 是的,没错,我刚刚发现。但它是如何工作的,因为我对这段代码的目标是: Optional users = userService.findById(employee.getId());应该是获取值在哪里获取employeeId。那么如何让这个employeeId包含在else if中。
  • @dotore 如何从以下位置获取此employeeId:可选 users = userService.findById(employee.getId());包含在 else if 中
  • 如果我理解正确,您想通过Employee.id 找到Users(不是Users 实例的id)。如果你没有找到它 => 返回一个异常。如果你找到它但Users.employeeId 不为空=> 返回异常"User employeID : "+ employee.getEmployeeId() +" Already Exist on Users"。这是正确的吗?

标签: java spring spring-boot entity field


【解决方案1】:

对于建议的解决方案,我将假设以下内容:

  • EmployeeUsers 之间存在关联。
  • 一个Employee只能与一个Users相关联
  • usernameUsers 的自然键
  • employeeIdEmployee 的自然键

实体

@Entity
public class Users {

  @Id
  // This one is an example, you can use the configuration you need
  @GeneratedValue(strategy = GenerationType.SEQUENCE, generator= "users_seq")
  @SequenceGenerator(name="users_seq", initialValue=1, allocationSize=1, sequenceName = "users_id_seq")
  private Long id;

  @Column(name = "fullname")
  private String fullName;

  // Probably this column should be unique and you need to configure in that way here and in your database
  @Column
  private String username;

  @Column
  private String password;

  // Getter & setter & constructors
}



@Entity
public class Employee {

  @Id
  // This one is an example, you can use the configuration you need
  @GeneratedValue(strategy = GenerationType.SEQUENCE, generator= "employee_seq")
  @SequenceGenerator(name="employee_seq", initialValue=1, allocationSize=1, sequenceName = "employee_id_seq")
  private Long id;

  /**
   * Assuming this is your specific identifier for an employee (not related with database PK)
   *    If the assumption is correct, this column should be unique and you need to configure in
   * that way here and in your database
   */
  @Column(name = "employeeid")
  private String employeeId;

  /**
   * Not sure if this relation could be nullable or not
   */
  @OneToOne
  @JoinColumn(name = "users_id")
  private Users users;

  // Getter & setter & constructors
}

如您所见,两个实体中都没有“重复列”,EmployeeUsers 之间存在单向 OneToOne 关系。如果您需要双向的,此链接将为您提供帮助:Bidirectional OneToOne

存储库

@Repository
public interface UsersRepository extends CrudRepository<Users, Long> {
  Optional<Users> findByUsername(String username);
}



@Repository
public interface EmployeeRepository extends CrudRepository<Employee, Long> {
  Optional<Employee> findByEmployeeId(String employeeId);
}

服务

@Service
public class UsersService {

  @Autowired
  private UsersRepository repository;

  public Optional<Users> findByUsername(String username) {
    return Optional.ofNullable(username)
            .flatMap(repository::findByUsername);
  }

  public Optional<Users> save(Users user) {
    return Optional.ofNullable(user)
            .map(repository::save);
  }
}



@Service
public class EmployeeService {

  @Autowired
  private EmployeeRepository repository;

  @Autowired
  private UsersService usersService;

  public Optional<Employee> insert(Employee newEmployee) {
    /**
     * The next line don't make sense:
     *
     *   Optional<Users> users = userService.findById(employee.getId());
     *
     * I mean:
     *
     *  1. Usually, id column is configured with @GeneratedValue and manage by database. So you don't need to ask
     *     if that value exists or not in Users.
     *
     *  2. Even if you are including id's values manually in both entities what should be "asked" is:
     *
     *    2.1 Is there any Users in database with the same username than newEmployee.users.username
     *    2.2 Is there any Employee in database with the same employeeId
     *
     *    Both ones, are the natural keys of your entities (and tables in database).
     */
    return Optional.ofNullable(newEmployee)
            .filter(newEmp -> null != newEmp.getUsers())
            .map(newEmp -> {
                isNewEmployeeValid(newEmp);

                // Required because newEmp.getUsers() is a new entity (taking into account the OneToOne relation)
                usersService.save(newEmp.getUsers());

                repository.save(newEmp);
                return newEmp;
            });
  }

  private void isNewEmployeeValid(Employee newEmployee) {
    if (usersService.findByUsername(newEmployee.getUsers().getUsername()).isPresent()) {
        throw new RuntimeException("Username: "+ newEmployee.getUsers().getUsername() +" exists in database");
    }
    if (repository.findByEmployeeId(newEmployee.getEmployeeId()).isPresent()) {
        throw new RuntimeException("EmployeeId: "+ newEmployee.getEmployeeId() +" exists in database");
    }
  }
}

【讨论】:

  • 谢谢先生您的示例...非常有帮助,然后我可以尝试这样做。我在您使用的样本中看到了@onetoone,这是我在尝试的。但是通过您的示例,我可以理解我的问题,然后我可以解决我的问题....非常感谢先生对此...
【解决方案2】:

读完 cmets 我已经明白你的问题了。

Users users = userService.findById(employee.getId()).orElseThrow(() -> new Exception("User ID : "+ employee.getId() +" Not Founded"));

现在您可以从返回的userService.findById(employee.getId()) 中获取您的employeeId 来自users

示例:

String employeeId = users.getEmployeeId(); // reference to your code

但在我看来,在这种情况下,您应该在usersemployee 之间建立关系@OneToOne,或者在employee 类中扩展users

One-To-One relation in JPA, hibernate-inheritance

【讨论】:

  • 好的,先生,我需要尝试您的解决方案。所以我需要将我的实体类更改为employeeId 成为表用户和员工之间的关系'@OneToOne。这是我目前的设计是错误的,因为我没有应用设计'@OneToOne 或在员工类中扩展用户......谢谢先生回答我的问题......我会在我尝试你的解决方案后更新先生.我希望这个解决方案能提供我一直在寻找的答案
  • 而且你不需要更新查询。 save() 也像更新一样工作。
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