JPA 命名查询 findAllBy 与 Long 和 String

Question

我对 spring data JPA 命名方法 findAllBy 有问题...这是我的实体：@Id@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "sequenceGenerator")@SequenceGenerator(name = "sequenceGenerator")private Long id;@Column(name = "entity_id")private Long entityId;@Column(name = "entity_name")private String entityName;@Column(name = "user_id")private Long userId;@Column(name = "rating")private Double rating;@Column(name = "like")private Long like;@Column(name = "dislike")private Long dislike;@Column(name = "review_title")private String reviewTitle;@Lob@Column(name = "review_comment")private String reviewComment;@Column(name = "time")private ZonedDateTime time;@ManyToOneprivate RatingType type;与 getter 和 setter。这是 ratingServiceImpl 中使用 @Autowired ratingRepository 的方法调用：List<Rating> ratings = ratingRepository.findAllByEntityIdAndEntityName(entityId, entityName);和存储库：@Repositorypublic interface RatingRepository extends JpaRepository<Rating, Long>,   JpaSpecificationExecutor<Rating> {     List<Rating> findAllByEntityIdAndEntityName(Long entityId, String entityName); }依赖关系：<dependency>    <groupId>org.springframework.boot</groupId>    <artifactId>spring-boot-starter-data-jpa</artifactId>    <version>1.5.10.RELEASE</version></dependency>

Answer 1

创建 JPA 实体时，尽量不要使用 DB 保留字作为列名和变量。

我相信问题出在

@Column(name = "like")
private Long like;

spring生成的SQL语句是：

...  RATING0_.LIKE[*] AS LIKE5_48_,

可以更改列名称吗？