【发布时间】:2018-04-11 13:47:39
【问题描述】:
我正在玩 Lombok 并且已经浏览了许多链接,但没有一个对我有用。
Person.java
@Setter @Getter
@ToString
@AllArgsConstructor
//@NoArgsConstructor
@RequiredArgsConstructor
@Entity
public class Person {
@Id
@GeneratedValue
private Long id;
@NotNull
@Size(min = 1, max = 20)
private String firstName;
@NotNull
@Size(min = 1, max = 50)
private String lastName;
}
PersonController.java
@RestController
@RequestMapping("/people")
public class PersonController {
@Autowired
private PersonRepository personRepository;
@RequestMapping(value = "", method = RequestMethod.POST)
@ResponseStatus(HttpStatus.CREATED)
public void createPerson(@RequestBody Person person) {
personRepository.save(new Person(person.getFirstName(), person.getLastName())); //line-34
}
}
但它不允许我创建两个参数的构造函数
Multiple markers at this line
- The constructor Person(String, String) is undefined
- The method save(S) in the type CrudRepository<Person,Long> is not applicable for the arguments
(Person)
第 34 行断线...
EDIT-1:
@RequestMapping(value = "/{id}", method = RequestMethod.PUT)
@ResponseStatus(HttpStatus.NO_CONTENT)
public void updatePerson(@PathVariable("id") Long id, @RequestBody Person person) {
Person existingPerson = personRepository.findOne(id);
existingPerson.setFirstName(person.getFirstName());
existingPerson.setLastName(person.getLastName());
personRepository.save(existingPerson);
}
这是错误
The method setFirstName(String) is undefined for the type Person
我所做的更改
@Setter @Getter
@ToString
@AllArgsConstructor
//@NoArgsConstructor
@RequiredArgsConstructor()
@Entity
public class Person {
@Id
@GeneratedValue
private Long id;
@NotNull
@Size(min = 1, max = 20)
private final String firstName;
@NotNull
@Size(min = 1, max = 50)
private final String lastName;
}
-====================
Edit-2
这是最终结果:
@Setter @Getter
@ToString
@AllArgsConstructor
@RequiredArgsConstructor
@Entity
public class Person {
@Id
@GeneratedValue
private Long id;
@NotNull
@Size(min = 1, max = 20)
private String firstName;
@NotNull
@Size(min = 1, max = 50)
private String lastName;
public Person(String firstName, String lastName){
this.firstName = firstName;
this.lastName = lastName;
}
}
【问题讨论】:
-
避免在 JPA 实体上使用 @EqualsAndHashCode。阅读此答案中的链接:stackoverflow.com/a/34299054/1331935
标签: java spring microservices lombok