Question

我有一个User类实体：

@Entity
@Table(name = "USERS")
@JsonIdentityInfo(generator = ObjectIdGenerators.IntSequenceGenerator.class, property = "@json-id")
public class User extends AbstractEntity {

  @Id
  @SequenceGenerator(name = "USERS_SEQ", sequenceName = "S_USER_ACCOUNT")
  @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "USERS_SEQ")
  @Column(name = "USER_ID", precision = 22, scale = 0)
  private Long userId;

  @Column(name = "ID_TYPE")
  private String idType;

  @Column(name = "COUNTRY_CODE", length = 3, nullable = false)
  @NotBlank
  private String countryCode;
  }
}

和类UserRepository从JpaRepository扩展

@Repository
public interface UserRepository extends JpaRepository<User, Long> {

  Page<User> findbyUser(User users, Pageable pageable);


}

当构建和编译源时，我有一条错误消息：

Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'userRepository': Invocation of init method failed; nested exception is org.springframework.data.mapping.PropertyReferenceException: No property findbyUser found for type User!

我想用实体中的任何字段搜索用户。你能说出你的想法吗？

Answer 1

谢谢大家。我刚刚研究了这个答案的解决方案。效果很好。

Example<User> example = Example.of(user); 
Page<User> pageResult = usersRepository.findAll(example, pageable);

Answer 2

您可以使用if(req.url == '/') { filename = '/public/index.html' } else { filename = path.basename(parsedURL.pathname); }注释按任何字段进行搜索，例如：

@Query

按countryCode搜索：

    @Query("SELECT user FROM User user WHERE (:userId IS NULL OR user.userId = :userId) " +
            "AND (:idType IS NULL OR user.idType = :idType) " +
            "AND (:countryCode IS NULL OR user.countryCode = :countryCode) ")
    Page<User> search(@Param("userId") Long userId, @Param("idType") String idType,
                                        @Param("countryCode") String countryCode, Pageable pageable);

按类型和countryCode搜索：

    userRepository.search(null, null, "US", new PageRequest(page, pageSize));

按ID搜索：

    userRepository.search(null, "someType", "US", new PageRequest(page, pageSize));

Answer 3

我建议你制作3种不同的方法：

userRepository.search(1L, null, null, new PageRequest(page, pageSize));

Page<User> findAllByUserId(Long userId, Pageable pageable);

Page<User> findAllByIdType(String idType, Pageable pageable);

在Page<User> findAllByCountryCode(String countryCode, Pageable pageable);中，获取所有网页的内容并将其合并为一个@Service

请注意，这些方法可以返回用户列表，而Page<User>可以返回@Controller。（这可以使Page<User>响应的构建更轻松，更容易）< / p>

JpaRepository for Java Spring：找不到类型User的findbyUser属性

3 个答案: