java - Spring Boot REST Hibernate - 创建用户

Question

我尝试在我的后端创建一个函数来创建一个用户，我使用 Spring Boot、Hibernate、JPA、PostgreSQL...这是我的代码:

用户.java

@Entity
@Table(name = "users")
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @NotNull
    @NotBlank
    @Size(max = 100)
    @Column(name = "firstName")
    private String name;

    @NotNull
    @NotBlank
    @Size(max = 30)
    @Column(name = "username", unique = true)
    private String username;

    @NotNull
    @NotBlank
    @Size(max = 150)
    @Column(name = "password")
    private String password;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "cityId", nullable = false)
    @JsonIgnore
    private City city;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "countryId", nullable = false)
    @JsonIgnore
    private Country country;

    // Getters and Setters
    ...
}

用户 Controller .java

@PostMapping("/users/{countryId}/{cityId}")
public User createUser(@PathParam(value = "countryId") Long countryId, @PathParam(value = "cityId") Long cityId,
        @Valid @RequestBody User user) {
    user.setCountry(countryRepository.findById(countryId)
            .orElseThrow(() -> new ResourceNotFoundException("Country not found with id " + countryId)));
    user.setCity(cityRepository.findById(cityId)
            .orElseThrow(() -> new ResourceNotFoundException("City not found with id " + cityId)));
    return userRepository.save(user);
}

UserRepository.java

@Repository
public interface UserRepository extends JpaRepository<User, Long> {
    List<User> findByCountryId(Long countryId);

    List<User> findByCityId(Long cityId);
}

我使用 Postman 进行测试。我尝试使用此 URL(1 = countryID，4 = cityId)和有效负载创建用户:

网址

localhost:8080/users/1/4

负载

{
    "name": "David",
    "username": "david",
    "password": "test",
}

我收到了这个错误...

错误:

{
    "timestamp": "2018-05-07T13:44:03.497+0000",
    "status": 500,
    "error": "Internal Server Error",
    "message": "The given id must not be null!; nested exception is java.lang.IllegalArgumentException: The given id must not be null!",
    "path": "/users/1/4"
}

2018-05-07 14:25:40.484 ERROR 17964 --- [io-8080-exec-10] o.a.c.c.C.[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.dao.InvalidDataAccessApiUsageException: The given id must not be null!; nested exception is java.lang.IllegalArgumentException: The given id must not be null!] with root cause

但是我不知道怎么解决这个问题

Answer 1

你应该在PostgreSQL中设置id列的数据类型为SERIAL

将 @PathParam 更改为 @PathVariable 应该有效。

您使用了错误的注释。

编辑

Spring 使用注解来做一些特殊的逻辑，从请求 URI 或请求体中提取值，并将它们映射到适当的注解参数。

您在参数上使用了错误的注释，因此未填充其值。

当您的存储库执行代码以针对null 查找时将抛出异常