SQLGrammarException：无法准备语句

Posted 2023-03-16

【中文标题】SQLGrammarException：无法准备语句

【英文标题】：SQLGrammarException: could not prepare statement

【发布时间】：2021-02-01 18:28:23

【问题描述】：

我有一个带有一个由 CrudRepository 寻址的 h2 数据库的 Spring Boot 应用程序。

public interface PetRepository extends CrudRepository<PetBE, Long> 

    @Query("SELECT p FROM PetBE p INNER JOIN TagBE t WHERE t.name IN :tags")
    public List<PetBE> findPetsByAnyTag(@Param("tags") List<String> tags);

我的实体如下

@Data
@NoArgsConstructor
@RequiredArgsConstructor
@AllArgsConstructor
@Entity
@Table(name = "pets")
public class PetBE 
    
    @Id
    private long id;

    @NonNull
    private String name;

    @ManyToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "category_id", referencedColumnName = "id")
    private CategoryBE category;
    
    @NonNull
    @ElementCollection
    private List<String> photoUrls;
    
    @ManyToMany(cascade = CascadeType.ALL)
    @JoinTable(
        name = "pet_tags", 
        joinColumns = @JoinColumn(name = "pet_id"), 
        inverseJoinColumns = @JoinColumn(name = "tag_id"))
    @OrderColumn(name = "id")
    private List<TagBE> tags;
    
    @Enumerated(EnumType.STRING)
    private StatusEnum status;

@Data
@NoArgsConstructor
@RequiredArgsConstructor
@Entity
@Table(name = "tags")
public class TagBE 

    @Id
    @NonNull
    private long id;
    
    @NonNull
    private String name;
    
    @ManyToMany
    private List<PetBE> pets;
    

但是当我调用我的方法时，我得到：

org.springframework.dao.InvalidDataAccessResourceUsageException: could not prepare statement; SQL [select petbe0_.id as id1_4_, petbe0_.category_id as category4_4_, petbe0_.name as name2_4_, petbe0_.status as status3_4_ from pets petbe0_ inner join tags tagbe1_ on where tagbe1_.name in (?)]; nested exception is org.hibernate.exception.SQLGrammarException: could not prepare statement
Caused by: org.h2.jdbc.JdbcSQLSyntaxErrorException: Syntax error in SQL statement "SELECT PETBE0_.ID AS ID1_4_, PETBE0_.CATEGORY_ID AS CATEGORY4_4_, PETBE0_.NAME AS NAME2_4_, PETBE0_.STATUS AS STATUS3_4_ FROM PETS PETBE0_ INNER JOIN TAGS TAGBE1_ ON WHERE[*] TAGBE1_.NAME IN (?)"; expected "NOT, EXISTS, INTERSECTS, UNIQUE"; SQL statement:
select petbe0_.id as id1_4_, petbe0_.category_id as category4_4_, petbe0_.name as name2_4_, petbe0_.status as status3_4_ from pets petbe0_ inner join tags tagbe1_ on where tagbe1_.name in (?) [42001-200]

我不知道从哪里开始，语法对我来说很好。 查询也会编译，错误仅在运行时出现。 我在这里错过了什么？

【问题讨论】：

logicbig.com/tutorials/java-ee-tutorial/jpa/…

【参考方案1】：

通过不使用定义的关系，您将引入交叉连接（但需要内部连接）。您可以在生成的查询的以下片段中看到它：

inner join tags tagbe1_ on where tagbe1_.name in (?)
                           ^

您可以通过使用第二个实体作为第一个实体的字段来解决此问题：

@Query("SELECT p FROM PetBE p INNER JOIN p.tags t WHERE t.name IN :tags")
public List<PetBE> findPetsByAnyTag(@Param("tags") List<String> tags);