如何解决如何在Java Spring上获取字段变量
我的学习项目遇到问题，例如“如果条件值为null，然后如果条件值字段为null”，例如我的代码紧随以下代码：
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....
}
，然后对于我的班级服务，我要插入带有仓库的数据Employee。如果在将数据插入表Employee之前有验证数据，我们需要检查表用户不为空，然后在字段 employeeId 为空。我的代码如下：
对于存储库 UserRepo.java 和 EmployeeRepo.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，而我尝试调试以查看employeeId的值，而我始终看到为Null，则其他问题。因此，对我的问题有任何想法，因为我尝试某些案例总是会因我的问题而失败。如果对我的问题有任何想法，请回答这些问题。非常感谢您回答我的问题。
解决方法
阅读评论后，我已经了解您的问题。
node.circle现在您可以从返回的Users users = userService.findById(employee.getId()).orElseThrow(() -> new Exception("User ID : "+ employee.getId() +" Not Founded"));
的{{1}}中获取employeeId；
示例：
users但是在我看来，在这种情况下，您应该在userService.findById(employee.getId())和String employeeId = users.getEmployeeId(); // reference to your code
之间建立@OneToOne的关系，或在users类中扩展employee。
One-To-One relation in JPA， hibernate-inheritance
,对于建议的解决方案，我将假设以下内容：
-
Employee和Users之间存在联系。 -
Employee只能与一个Users关联 -
username是Users的自然键 -
employeeId是Employee的自然键
所以实体：
@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,generator= "employee_seq")
@SequenceGenerator(name="employee_seq",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“重复的列”，并且在OneToOne和Employee之间存在单向Users 关系。如果您需要双向链接，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");
}
}
}
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