如何加入jpql？ - org.hibernate.query.SyntaxException

Question

拥有这些实体：

用户.java：

@Getter
@Setter
@NoArgsConstructor
@AllArgsConstructor
@Entity
@Table(name = "user")
public class User {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Long id;
    private String username;
    @OneToMany(mappedBy = "user", fetch = FetchType.EAGER)
    private List<Nickname> nicknames;
}

昵称.java:

@Getter
@Setter
@NoArgsConstructor
@AllArgsConstructor
@Entity
@Table(name = "nickname")
public class Nickname {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    private String value;
    @ManyToOne
    @JoinColumn(name = "user_id")
    private User user;

    @Override
    public String toString() {
        return "Nickname{" +
                "id=" + id +
                ", value='" + value + '\'' +
                '}';
    }
}

和昵称存储库：

public interface NicknameRepository extends JpaRepository<Nickname, Long> {
    @Transactional
    @Modifying
//    @Query("DELETE FROM Nickname n WHERE n.value = :value AND n.user.username = :username")
    @Query("DELETE FROM Nickname n JOIN n.user u where n.value = :value AND u.username = :username")
    void deleteNicknameByValue(@Param("username") String username, @Param("value") String value);
}

我正在尝试加入这两个实体，因为我需要来自

username

的

User

和来自

value

的

Nickname

。我明白了

java.lang.IllegalArgumentException: org.hibernate.query.SyntaxException: At 1:23 and token 'JOIN', mismatched input 'JOIN', expecting one of the following tokens: <EOF>, WHERE [DELETE FROM Nickname n JOIN n.user u where n.value = :value AND u.username = :username]

这个连接有什么问题吗？

Answer 1

来自文档

相对于 SELECT 查询的结构，DELETE 查询的结构非常简单。 DELETE 查询不能包含多个变量和 JOIN，也不能包含 GROUP BY、HAVING 和 ORDER BY 子句。

此处可以使用

IN

子句来删除

Nickname

。

@Query("DELETE FROM Nickname n where n.value = :value AND n.id IN (SELECT n2.id FROM Nickname n2 JOIN n2.user u WHERE u.username = :username)")
void deleteNicknameByValue(@Param("username") String username, @Param("value") String value);