JPA named query findAllBy with Long and String

Question

i have problem with spring data JPA naming method findAllBy...

This is my entity :

@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;

@ManyToOne
private RatingType type;

with getters and setters.

This is method call in ratingServiceImpl with @Autowired ratingRepository :

List<Rating> ratings = ratingRepository.findAllByEntityIdAndEntityName(entityId, entityName);

and repository :

@Repository
public interface RatingRepository extends JpaRepository<Rating, Long>, 
  JpaSpecificationExecutor<Rating> {
     List<Rating> findAllByEntityIdAndEntityName(Long entityId, String entityName); 
}

Dependency :

<dependency>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-starter-data-jpa</artifactId>
    <version>1.5.10.RELEASE</version>
</dependency>

Caused by :

Caused by: org.h2.jdbc.JdbcSQLException: Syntax error in SQL statement "SELECT RATING0_.ID AS ID1_48_, RATING0_.DISLIKE AS DISLIKE2_48_, RATING0_.ENTITY_ID AS ENTITY_I3_48_, RATING0_.ENTITY_NAME AS ENTITY_N4_48_, RATING0_.LIKE[*] AS LIKE5_48_, RATING0_.RATING AS RATING6_48_, RATING0_.REVIEW_COMMENT AS REVIEW_C7_48_, RATING0_.REVIEW_TITLE AS REVIEW_T8_48_, RATING0_.TIME AS TIME9_48_, RATING0_.TYPE_ID AS TYPE_ID11_48_, RATING0_.USER_ID AS USER_ID10_48_ FROM RATING RATING0_ WHERE RATING0_.ENTITY_ID=? AND RATING0_.ENTITY_NAME=? "; expected "identifier"; SQL statement: select rating0_.id as id1_48_, rating0_.dislike as dislike2_48_, rating0_.entity_id as entity_i3_48_, rating0_.entity_name as entity_n4_48_, rating0_.like as like5_48_, rating0_.rating as rating6_48_, rating0_.review_comment as review_c7_48_, rating0_.review_title as review_t8_48_, rating0_.time as time9_48_, rating0_.type_id as type_id11_48_, rating0_.user_id as user_id10_48_ from rating rating0_ where rating0_.entity_id=? and rating0_.entity_name=? [42001-196]

other locations of exception on this link

Thank you for solution in advance.

Possible duplicate of [hibernate h2 embeddable list expected "identifier"](https://stackoverflow.com/questions/46737430/hibernate-h2-embeddable-list-expected-identifier) — Carlos López Marí, Oct 07 '19 at 14:59
Better to change `@Column(name = "like")` by `@Column(name = "SOME_OTHER_NAME_BECAUSE_\`LIKE\`_IS_AN_RESERVED_KEY_WORD")` — Zorglube, Oct 07 '19 at 15:22
If you just copy paste the query in the db you would have understood the issue. Probably reserved keywords. — Karthik R, Oct 07 '19 at 16:54

Ioannis Barakos · Accepted Answer · 2019-10-07T16:14:56.083

1

Try not to use DB reserved words as column names and variables when creating your JPA Entity.

I believe the problem is in

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

as the SQL statement produced by spring is:

...  RATING0_.LIKE[*] AS LIKE5_48_,

Can you change the column name?

edited Oct 07 '19 at 16:14

answered Oct 07 '19 at 15:06

Ioannis Barakos

1,319
1
11
16

You can try `@Column(name = "'like'") private Long like;` or `@Column(name = "\`like\`") private Long like;` – Zorglube Oct 07 '19 at 15:19

JPA named query findAllBy with Long and String

1 Answers1