Hibernate 5.2.17：ORA-01797：此运算符后必须跟有 ANY 或 ALL

Question

我有一个 Oracle 18.4.0 XE 数据库，我正尝试从 JPA 2.1 访问它，该数据库由 Hibernate 5.2.17 实现。

我在 2 个实体之间有 ManyToMany 连接：

public class PermissionEntity implements Serializable {
    private static final long serialVersionUID = -3862680194592486778L;

    @Id
    @GeneratedValue
    private Long id;

    @Column(unique = true)
    private String permission;

    @ManyToMany
    private List<RoleEntity> roles;
}

public class RoleEntity implements Serializable {
    private static final long serialVersionUID = 8037069621015090165L;

    @Column(unique = true)
    private String name;

    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private Long id;

    @ManyToMany(fetch = FetchType.LAZY, mappedBy = "roles")
    private List<PermissionEntity> permissions;
}

当尝试运行 PermissionRepository 上的 Spring Data JPA 请求时：findAllByPermission(Iterable<String> permissions)，出现以下异常：

Error : 1797, Position : 140, Sql = select permission0_.id as id1_0_, permission0_.permission as permission2_0_ from PermissionEntity permission0_ where permission0_.permission=(:1  , :2 ), OriginalSql = select permission0_.id as id1_0_, permission0_.permission as permission2_0_ from PermissionEntity permission0_ where permission0_.permission=(? , ?), Error Msg = ORA-01797: this operator must be followed by ANY or ALL

Answer 1

使用'in'关键字：findAllByPermissionIn(Iterable<String> permissions).

这会产生这样的查询：where permission0_.permission IN (:permissions)。

Answer 2

您要让 Spring 数据 Jpa 引擎搜索 Permission，其中权限等于列表。它应该使用 IN 运算符，因此您的方法名称应该是：

findAByPermissionIn(Iterable<String> permissions)

Hibernate 5.2.17：ORA-01797：此运算符后必须跟有 ANY 或 ALL

Hibernate 5.2.17: ORA-01797: this operator must be followed by ANY or ALL

java

spring

hibernate

spring-data-jpa

oracle18c